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\leftline{\font\title=cmr17\title Tsit-Yuen Lam. Representation theory.}
\bigskip
\leftline{Notes by Dmitri Pavlov, pavlov@math.}
\section Introduction.
All representations are finite-dimensional.
Most groups will be finite.
We assume that $k$ is a commutative ring (later: field).
\proclaim Classic Definition.
An $n$-dimensional representation of a group~$G$ is a group homomorphism $D\colon G\to\GL_n(k)$.
(D comes from Darstellung.)
Two representations $D$~and~$D'$ are said to be equivalent if there is a $T\in\GL_n(k)$ such that
for all $g\in G$ we have $T^{-1}D(g)T=D'(g)$.
The character of $D$ is a function $\chi_D\colon G\to k$ such that $\chi_D(g)=\tr(D(g))$ for all $g\in G$.
\proclaim Note. If two representations are equivalent, they have the same characters.
If two elements of the group are conjugate ($g\sim g'$), then $\chi_D(g)=\chi_D(g')$.
We call $\chi_D$ a ``class function'' because it is constant on each conjugacy class.
\proclaim Definition. A representation $D$ is faithful if the corresponding homomorphism is injective.
\proclaim Modern Definition. (80 years old.)
A $G$-module (over $k$) is a $k$-module (usually finitely generated)
with a $k$-linear $G$-action.
(Note: $g$ must act as a $k$-automorphism because $g^{-1}$ exists.)
Isomorphisms of $G$-modules make obvious sense.
We say that $G$ acts faithfully if the only element that acts as identity is the unit.
To link these definitions we do the following. Suppose that we have a homomorphism
$D\colon G\to\GL_n(k)$. We set $g\cdot v=D(g)(v)$.
Conversely, if we have a $G$-module, then we define $D(g)(v)=g\cdot v$.
\proclaim Propositions. Two representations $D$ and $D'$ are equivalent iff
two associate $G$-modules are equivalent.
\section Ring theoretic perspective.
Form a group ring $kG=k[G]$, which is a $k$-algebra.
\proclaim Observation. $G$-modules over~$k$ are the same as $kG$-modules.
\proclaim Definition. Representation afforded by a $G$-module~$V$ over~$k$
is irreducible iff $V\ne0$ and $V$ has no non-trivial $kG$-submodules.
Representation afforded by a $G$-module~$V$ over~$k$
is indecomposable iff $V\ne0$ and $V\ne V_1\oplus V_2$ for non-trivial $kG$-modules $V_1$~and~$V_2$.
An irreducible representation is indecomposable but not vice versa.
One dimensional representations are always irreducible and indecomposable.
\proclaim Example. Let $k=\F_2$, let $G$ be a cyclic group of order~2 generated by element~$\sigma$,
and let $V=kG$ with a left action.
This module is not irreducible because $\{0,\sigma\}$ is a non-trivial $kG$-submodule.
This module is indecomposable because $\{0,\sigma\}$ is the only non-trivial $kG$-submodule.
\section Matricial perspective.
Suppose that $D$ is a reducible representation with $V_0\subset V$ being a non-trivial
$kG$-invariant submodule.
Choose a basis of $V_0$ and supplement it to a basis of $V$.
We have two representations: $g\to D_1(g)$ afforded by $V_0$ and $g\to D_2(g)$ afforded by $V/V_0$.
Then the matrix corresponding to $D(g)$ has the form $\pmatrix{D_1(g)&E(g)\cr0&D_2(g)\cr}$.
If $D$ is decomposable, then $E(g)=0$.
\proclaim Note. One-dimensional representation is a homomorphism $G\to k^*=\GL_1(k)$.
Two such representations are equivalent iff they are equal.
\section Composition factors.
If $V$ is a $kG$-module then there exists a composition series $0=V_0\subset V_1\subset\cdots\subset V_m=V$,
with all $V_k$ different. Here $V_{k+1}/V_k$ are simple $kG$-modules.
The sequence of these composition factors is unique up to a permutation.
In our earlier example the composition factors are trivial one-dimensional representations.
\section Direct sums.
We have $\chi_{V\oplus V'}=\chi_V+\chi_{V'}$.
Scalar extensions. If we have an extension of fields $K/k$ and a representation~$V$ over~$k$,
then we have a representation over~$K$, which satisfies the equation $V^K=K\otimes_k V$.
The character of a representation is the sum of the characters of its composition factors.
If we recall our example, we can easily see that the equivalence class of $kG$-module
is not determined by its character, because the character of the module in the example is~0
and the character of the direct sum of two copies of~$k$ is also~0.
\proclaim Example. If $G$ is a finite group, then we have a left regular representation.
Its character~$\chi$ satisfies the following equations: $\chi(1)=|G|$ and $\chi(g)=0$ for $g\ne1$.
\proclaim Review of simple modules.
Let $R$ be any ring. An $R$-module~$V$ is simple iff $V\ne0$ and every element of~$V$
generates~$V$.
An $R$-module~$V$ is simple iff it is isomorphic to $R/m$ for some maximum left ideal~$m$ of~$R$.
Here $m$ is uniquely determined if $R$ is commutative.
If $R$ is a finite-dimensional $k$-algebra ($k$ is a field), then there are
only finitely many simple $R$-modules up to isomorphism.
Proof: Look at the left regular module of~$R$. By Jordan-H\"older theorem
we have finitely many composition factors $V_1$, \dots, $V_n$.
If $V$ is simple, we have $V=R/m$. We can complete the two-element series $m$ and $R$ to a Jordan-H\"older
series.
\proclaim Master Theorem. (To be proved.)
If $G$ is a finite group with $r$ conjugacy classes and $k$ is the field of complex numbers,
then the number of irreducible representations is equal to~$r$.
If $n_i$ is the dimension of $i$th irreducible representation of~$G$, then it divides~$|G|$.
Also $\sum_kn_k^2=|G|$ (magic equation).
Every finite-dimensional $kG$-module is uniquely a direct sum of irreducible representations.
(The direct sum itself is not unique.)
\proclaim Example. Let $G=\langle\sigma,\phi\mid\sigma^7=\phi^3=1,\phi^{-1}\sigma\phi=\sigma^2\rangle$.
We have $|G|=21$.
We have 3 one-dimensional representations of~$G$.
Representatives of conjugacy classes: 1, $\sigma$, $\sigma^3$, $\phi$, $\phi^2$.
By magic equation we discover that there are only two more irreducible representations, which
are 3-dimensional.
For finite abelian group we get a character table in the following way: let $G^*=\Hom(G,\C^*)$ be
the character group of~$G$. The character table consists of these characters line by line.
We have non-canonical isomorphism between $G$~and~$G^*$ and canonical isomorphism
between $G$~and~$G^{**}$.
All irreducible representations have dimension~1.
Another perspective (without Master Theorem).
\proclaim Theorem. Let $G$ be a finite abelian group, $k$ be algebraically closed field.
Then any simple $kG$-module $V$ is 1-dimensional.
\proclaim Proof. It suffices to show that for any $g\in G$ the operator $D(g)$ is a scalar multiplication.
Let $\lambda\in k$ be an eigenvalue of $D(g)$ and $E(\lambda)$ be the $\lambda$-eigenspace
of~$D(g)$. Obviously, it is invariant under $G$-action. Therefore, $D(g)=\lambda I_n$.
\proclaim Remark. We need only assume that the polynomial $x^e-1$ splits over~$k$, where
$e$ is the exponent of~$G$.
But if $k$ is arbitrary, the theorem does not work.
The cyclic group of order~3 acts irreducible on~$\Q^2$ by rotations by $2\pi/3$.
\proclaim Example. Character table of quaternion group (of order 8).
We use a complex matrix model of~$\H$.
It is easy to see that $D(1)$, $D(i)$, $D(j)$, $D(k)$ are linearly independent.
Therefore, $\C\otimes_\R\H$ is isomorphic to $\Mat_2(\C)$.
Restriction to $G$ gives us a 2-dimensional complex representation.
It is irreducible, since $D(G)$ $\C$-spans $\Mat_2(\C)$.
We also have four obvious 1-dimensional representations.
\proclaim A construction idea.
Denote by $k$ be a field and by $D$ a division algebra over~$k$.
Denote by $G$ a subgroup of $D^*$ such that $G$ spans~$D$ over~$k$.
Then $D$ is a simple $kG$-module.
\proclaim Proof.
Trivial.
\proclaim Example. Denote by $G$ the cyclic group of order~$n$. Construct all simple $\Q G$-modules.
\proclaim Solution. Denote by $d$ a divisor of~$n$.
Take $V_d$. By previous theorem, adjoining a primitive $d$th root of~1 to~$\Q$
we obtain a simple $\Q G$-module.
We have $\dim_\Q V_d=\varphi(d)$.
Composition factors of $\Q G$ regarded as a left module include all $V_d$.
Since the sum of their dimensions is equal to~$n$, we have listed all simple modules over~$\Q G$.
\proclaim Example. Now we want to find all simple $\R G$-modules, where $G$ is the quaternion group.
\proclaim Solution. At dimension~1 we have 4 $\R G$-representations.
Now denote by $\H$ the real quaternions. They form a simple $\R G$-module.
If we tensor multiply this by~$\C$, it splits into two irreducible modules.
The multiplicative group of quaternions has other finite subgroups
apart from the classical quaternion group.
For example, it contains generalized quaternion group of order~$4m$.
Take inside $\C^*$ a cyclic subgroup of order~$2m$ and adjoin~$j$ to it.
We also have other subgroups: binary tetrahedral group (order 24), binary octahedron group
(order 48), binary icosahedral group (order 120).
Hurwitz defined the ring of integral quaternions inside rational quaternions.
It consists of all quaternions with integer coefficients and with simultaneously semi-integer
coefficients.
We want to obtain all irreducible complex representations of tetrahedron group~$A_4$
and binary tetrahedron group~BT.
The conjugacy classes of $A_4$ are 1 (1), $\tau$ (3), $\sigma$ (4), $\sigma^2$ (4),
where $\tau=(12)(34)$ and $\sigma=(123)$.
The character table is
$$\matrix{
&1&\tau&\sigma&\sigma^2\cr
&1&3&4&4\cr
\chi_1&1&1&1&1\cr
\chi_2&1&1&\omega&\omega^2\cr
\chi_3&1&1&\omega^2&\omega\cr
\chi_4&3&-1&0&0\cr
}$$
The first three characters are 1-dimensional representations.
The remaining representation must have dimension~3.
It comes from representation of~$A_4$ as group of symmetries of tetrahedron.
\proclaim Theorem. Binary tetrahedron group is isomorphic to $\SL_2(\F_3)$.
\proclaim Proof. We sketch two different approaches.
First, easy counting shows that $\SL_2(\F_3)$ has 24 elements.
Recall that ${\rm BT}_{24}$ is a semidirect product of $Q_8$~and~$C_3$,
we just find the same structure inside
$\SL_2(\F_3)$.
We write down a unique 2-Sylow subgroup of~$\SL_2(\F_3)$.
Then we write down a matrix~$\sigma_0$ of order~3
and compute the conjugation action of it on 2-Sylow subgroup.
Another way to prove this fact is as follows:
Recall 2-dimensional irreducible complex representation:
$D(i)=\pmatrix{i&0\cr0&-i\cr}$ and $D(j)=\pmatrix{0&1\cr-1&0}$.
The rest is left to reader as an exercise.
\proclaim Dihedral group.
$D_n=\langle r,s\mid r^n=s^2=1{\rm\ and\ }srs^{-1}=r^{-1}\rangle$.
If $n$ is odd, then $[G,G]=\langle r\rangle$ and $G/[G,G]=\langle s\rangle$. We have two
1-dimensional representations.
If $n$ is even, then $[G,G]=\langle r^2\rangle$ and $G/[G,G]=\langle r\rangle\times\langle s\rangle$.
We have four 1-dimensional representations.
For $n=2$ we have one 2-dimensional representation coming from symmetries of $n$-gon on the plane.
We now see that $Q_8$ and $D_4$ have the same character tables but are not isomorphic.
For higher $n$ we get the following results:
\proclaim Future theorem.
If $\Char k$ does not divide $n!$, then the reduced representation is irreducible over~$k$.
Suppose $E$ is a finite $G$-set. We can easily construct a $G$-module with $E$ as a basis.
This module is reducible, since elements with equal coefficients form a nontrivial
submodule. Denote by $\bar V$ the corresponding factor module.
We have $\chi_{\bar V}(g)=|{\rm Fix}_E(g)|-1$.
We will show that if $\Char k$ does not divide $n!$ and $G$ is a symmetric group,
then the reduced module is simple.
We can use this fact to understand representation of $S_3$ and $S_4$.
We work out all irreducible complex representations of~$S_4$.
Two of them are trivial one-dimensional representations.
Inside $S_4$ we have normal Klein 4-group. Its factor group is $S_3$, therefore,
one two-dimensional representation is obtained via pullback from $S_3$ representation.
Two 3-dimensional representations come from tetrahedron and octahedron.
\section Chapter 1.
Notational Convention: Write homs of modules on the opposite side of scalars.
Homo-law: $(rm)f=r(mf)$.
If we let $E=\End({}_RM)$, then $M={}_RM_E$ is $(R,E)$-bimodule.
We have $\End({}_RR)=R$: $r\to[x\to xr]$.
Similarly, $\End(R_R)=R$: $r\to[x\to rx]$.
Let $E=\End({}_RM)$. Then $M^n=nM=\oplus_{n\rm\ copies}M$.
We have $\End(M^n)=\Mat_n(E)$.
Schur's lemma: endomorphism ring of a simple module is a division ring.
A module $M$ is called semisimple iff every submodule splits iff it is a (direct) sum of simple submodules
iff it is a (direct) sum of all of its simple submodules.
A semisimple module~$M$ is indecomposable iff it is simple.
Semisimple modules are closed under arbitrary direct sums, submodules and quotient modules.
If $M$ is semisimple, then $M$ is finitely generated iff it has finite length iff it is
a simple sum of simple modules.
In this case, $M$ is a direct sum of its composition factors.
A ring $R$ is called left-semisimple iff $_RR$ is semisimple.
A ring is semisimple iff all $R$-modules are semisimple.
A ring $R$ is called simple iff $R\ne0$ and the only (two-sided) ideals are 0~and~$R$.
(Warning: Semisimple rings are not simple in general.)
Ideals of a ring~$R$ are in bijective correspondence with ideals of the ring~$\Mat_n(R)$.
Therefore, if $R$ is simple, then so is $\Mat_n(R)$.
In particular, matrices over division ring form simple ring.
Moreover, matrices over division ring form semisimple ring,
more precisely: $\Mat_n(D)=nD^n$ as left modules.
Therefore, $D^n$ is the only simple $\Mat_n(D)$-module.
Now note that $\End({}_{\Mat_n(D)}D^n)=D$.
\proclaim Artin-Wedderburn theorem.
A ring $R$ is left semisimple iff it is isomorphic to finite direct product of matrix
rings over division ring.
\proclaim Proof. Write $_RR$ as a direct sum of simple modules $\oplus_kn_kM_k$.
Now we see that $\End(\oplus_kn_kM_k)=\prod_k\Mat_{n_k}(\End(M_k))$.
\proclaim Definition. A bimodule $_SV_T$ is faithfully balanced if the ring homomorphisms
$S\to\End(V_T)$ and $T\to\End({}_SV)$ are both isomorphisms.
Omnipresence: Start with any module $_SV$. (We may replace $S$ by its image in $\End V$.
Now $S$ acts faithfully.) Now let $T=\End({}_SV)$, so that $V={}_SV_T$.
Then replace $S$ by $\End(V_T)$.
Now this bimodule is faithfully balanced.
\proclaim Important example. For any ring~$R$ the ring $_{\Mat_n(R)}R^n_R$ is faithfully
balanced bimodule.
This includes for $n=1$ the case $_RR_R$.
Nice observation: $_{\Mat_n(R)}R^n$ is simple iff $n\ne0$ and $R$ is a division ring.
If $R$ is division ring, the statement is trivial.
To prove the other implication we use Schur's lemma.
\proclaim Artin-Wedderburn theorem. A ring $R$ is a left semisimple ring iff $R$ is
a finite direct product of matrix rings over division rings: $\Mat_{n_i}(D_i)$.
\proclaim Proof of uniqueness. It suffices to describe $n_i$ and $D_i$ in terms of the ring~$R$.
Note that $D_i^{n_i}$ is a simple $R$-module if we let other components of~$R$ act trivially.
Therefore, the number of rings in the decomposition is equal to the number of simple left $R$-modules.
$D_i$ is the endomorphism ring of $i$th simple module.
Now $n_i$ is the dimension of this module over~$D_i$.
Also $n_i$ is the multiplicity of $V_i$ as composition factor.
Moreover, $n_i$ is the matrix size of the $i$th component in the Artin-Wedderburn decomposition.
\proclaim Corollary. A left semisimple ring is also right semisimple ring and vice versa.
Also a left semisimple ring is Artinian, and, therefore, Noetherian.
\proclaim Corollary. A semisimple ring is a direct sum of its indecomposable ideals,
which are uniquely determined up to a permutation.
\proclaim Corollary. A commutative ring is semisimple iff it is isomorphic to a finite direct
product of fields.
Now we want to relate simple rings to semisimple.
\proclaim Theorem. If $R$ is a simple ring, then $R$ is semisimple iff $R=\Mat_n(D)$ where $D$
is a division ring iff $R$ is left artinian.
\proclaim Proof. We only need to prove that an artinian simple ring is semisimple.
Take a left minimal ideal~$I$ of~$R$.
Now take let $B$ be the sum of all left ideals that are isomorphic to this one.
We easily see that this sum is also a right ideal,
therefore it coincides with the whole ring.
By artinity we obtain the desired result.
Note that there is no simple classification of left Noetherian rings.
\proclaim Maschke's theorem (1899).
Suppose $k$ is a field and $G$ is a group. Then $kG$ is semisimple iff the
characteristic of~$k$ does not divide~$|G|$ and $G$ is finite.
If $kG$ is semisimple, we call this case ordinary. Otherwise we call it modular.
\proclaim Modern proof. In the ordinary case we need to prove that every
exact sequence of $kG$-modules splits.
Fix a $k$-homomorphism $\lambda\colon V\to W$ such that $\lambda$ is identity on~$W$.
Now average $\lambda$ over all elements of $G$, obtaining
a $kG$-homomorphism.
Assume that $kG$ is semisimple. $G$ may be infinite.
Consider the augmentation map $\epsilon\colon kG\to k$,
$\epsilon(\sum a_gg)=\sum_{g\in G}a_g$.
Now take the kernel of $\epsilon$.
This kernel splits. Denote by $J$ its complement, which is a left ideal.
For any $\alpha\in J$ such that $\alpha\ne0$ we have $(g-1)\alpha\in(\ker\epsilon)\cap J=0$.
Therefore, $\alpha=g\alpha$ for all $g$, hence $\alpha=a\sum_{h\in G}h$.
Moreover, $G$ is finite.
Now note that $\epsilon(\alpha)=a|G|\ne0$, therefore $\Char k$ does not divide~$|G|$.
Maschke's original approach involved hermitean forms and orthogonal complements.
In real case replace the hermitean product by ordinary inner product.
\proclaim Proposition. Every complex representation is equivalent to a unitary representation.
Every real representation is equivalent to a orthogonal representation.
\proclaim Proof. Average an arbitrary hermitean form over all elements of group.
\proclaim Corollary. Every real or complex $G$-submodule splits.
\proclaim Proof. Take an orthogonal complement.
From now on we assume that $\Char k$ does not divide $|G|$.
Denote by $M_i=D_i^{n_i}$ the simple $G$-modules.
We have $D_i=\End_R(M_i)$. Let $m_i=\dim_kM_i$ and $d_i=\dim_kD_i$.
We have $m_i=n_id_i$, the Wedderburn components of $kG$ are $\End(M_i)$.
We have $_RR=\oplus_kn_kM_k$.
Now we obtain the general magic equation $|G|=\sum_in_im_i=\sum_in_i^2d_i\le\sum_im_i^2$.
If $k$ is algebraically closed, then $d_i=1$ and we obtain the usual magic equation.
The number $r$ is equal to the number of conjugacy classes.
To prove this note that the number of conjugacy classes is equal to the
dimension of the center of $kG$.
\proclaim Example.
Suppose $G$ is an abelian group.
We have $n_i=1$ and every $D_i$ is a field, therefore $kG$ is a direct product of fields.
Every $D_i$ supports a simple $kG$-module.
If $k$ has all necessary roots of unity, then $D_i=k$.
Therefore, if $G$~and~$H$ are abelian and $|G|=|H|$, then $\C G$ is isomorphic to $\C H$.
\proclaim Example. Construct the Wedderburn decomposition of $\Q G$, where $G$ is a cyclic group
of order~$n$.
We have $\Q G=\Q[t]/(t^n-1)$.
By Chinese remainder theorem we have $\Q G=\prod_{d\divides n}Q[t]/(\Phi_d(t))=
\prod_{d\divides n}\Q(\zeta_d)$. The last product is the Wedderburn decomposition.
Now we replace $\Q$ by $\R$. We have two cases. In the first case the order of group is even.
In this case it is easy to see that the Wedderburn decomposition is $\R\times\R\times\C^{n/2-1}$.
In the remaining case the Wedderburn decomposition is $\R\times\C^{(n-1)/2}$.
\halign{
&$#$\hfil\cr
G&\Q G&\R G&\C G\cr
C_{12}&\Q^2\times\Q(\omega)^2\times\Q(i)\times\Q(\zeta_{12})&\R^2\times\C^5&\C^{12}\cr
S_3&\Q^2\times\Mat_2(\Q)&\R^2\times\Mat_2(\R)&\C\times\C\times\Mat_2(\C)\cr
Q_8&\Q^4\times\H_\Q&\R^4\times\H&\C^4\times\Mat_2(\C)\cr
D_4&\Q^4\times\Mat_2(\Q)&\R^4\times\Mat_2(\R)&\C^4\times\Mat_2(\C)\cr
A_4&\Q\times\Q(\omega)\times\Mat_3(\Q(\sqrt{-7}))&\R\times\C\times\Mat_3(\R)&\C^3\times\Mat_3(\C)\cr
\langle\phi,\sigma\rangle&\Q\times\Q(\omega)\times&&\C^3\times\Mat_3(\C)^2\cr
S_4&&&\C^2\times\Mat_2(\C)\times\Mat_3(\C)^2\cr
}
\section Theorem of splitting fields.
How simple modules behave under extension of scalars?
%Problem VI.8 Show that an artinian module is semisimple iff every simple submodule splits.
\proclaim Theorem. A left artinian ring $R$ is simple iff there exists a faithful simple left $R$-module $M$.
\proclaim Proof. (Jacobson radical argument.)
Assume that there is a faithful simple left module~$M$ over left artinian ring~$R$.
First we show that $R$ is semisimple.
Check that $_RR$ is a semisimple module. It suffices to show
that any minimal left ideal $I\subset{}_RR$ splits. Fix $0\ne a\in I$.
Then $am\ne0$ for some $m\in M$.
Hence $Ram=M$. In particular, $m=ram$ and $(1-ra)m=0$, therefore $R(1-ra)$ is a proper left
ideal of~$R$. Denote by $J$ a maximal left ideal that contains~$R(1-ra)$.
Note that $I\cap J=0$, therefore $_RR=I\oplus J$.
\proclaim Burnside's theorem. Denote by $D$ a skew field.
If we take a ring~$R$ we have $D\subset R\subset\Mat_n(D)$.
Let $M=D^n$. Then $R=\Mat_n(D)$ iff $_RM$ is simple and $\End({}_RM))=D$.
\proclaim Proof. Certainly, $R$ is a left artinian ring. Moreover, $M$ is simple and faithful.
By the theorem above, $R$ is simple. The Wedderburn theory applies to $R$, hence $R=\Mat_n(D)$.
\proclaim Homomorphism theorem. Let $R$ be a $k$-algebra (not necessarily finite-dimensional)
and $K\supset k$ be a field extension. Let $M$~and~$N$ be left $R$-modules.
Then the natural map $\Hom_R(M,N)\to\Hom_{R^K}(M^K,N^K)$ is a $K$-vector space isomorphism.
Here $R^K=K\otimes_kR$ and $M^K=K\otimes_kM$.
\proclaim Proof. First Course, page~104.
\proclaim Theorem. Let $R$ be a $k$-algebra (not necessarily finite-dimensional).
Let $_RM$ be a simple $R$-module. We have $\End({}_RM)=k$ iff $R\to\End(M_k)$ is surjective
iff for any field extension $K\supset k$ the module $M^K$ is a simple $R^K$-module
iff there is an algebraically closed field extension $E\supset k$ such that $M^E$ is
a simple $R^E$ module.
\proclaim Proof. (2) follows from (1) applied to image of $R\to\End(M_k)$.
(3) follows from (2) because we have surjective morphism $R^K\to(\End(M_k))^K=\End_K(M^K)$,
hence $M^K$ is $R^K$-simple.
(1) follows from (4): $\End_{R^E}(M^E)=E$ is isomorphic to $(\End_R(M))^E$, therefore
$\End({}_RM)=k$.
\proclaim Definition. Let $R$ be a finite-dimensional $k$-algebra. Field extension
$K\supset k$ is a splitting field for~$R$ if every simple $R^K$-module is absolutely simple.
Absolutely simple means that its isomorphism ring is the ring of scalars.
Alternatively, it stays simple under any extension.
Algebraic closure is an example of splitting field.
A simple module over semisimple ring is absolutely simple if its Wedderburn component
looks like $\Mat_n(k)$.
\proclaim Proposition 1. If $k\subset K$ is a field extension, $R$ is a finite-dimensional
$k$-algebra, then any simple $R^K$-module~$U$ is a composition factor of~$V^K$
for some simple $R$-module~$V$.
\proclaim Proof. Insert $U$ into composition series of $_RR^K$.
\proclaim Proposition 2. Splitting field represents ``stable state''.
If $k\subset K\subset L$ are field extensions, $K$ is a splitting field for $k$-algebra~$R$
and $U_i$ is a complete set of $R^K$ simple modules,
then $U_i^L$ is a complete set of $R^L$ simple modules. And $L$ is a splitting field for~$R$.
Moreover, assume that $L$ is a splitting field. Then $K$ is a splitting field
iff every simple $R^L$-module is defined over $K$ (is isomorphic to tensor product of $L$
and some $R^K$-simple module.
\proclaim Theorem. For any finite-dimensional $k$-algebra there is a finite
field extension $K/k$ that splits~$R$.
\proclaim Proof.
Take $L=\bar k$ and construct~$K$.
Take maximal left ideals $A_i$ in $R^L$. Then $M_i=R^L/A_i$ gives a complete set of $R^L$-simples.
Take a big finite extension $K$ within $L$ such that $A_i\cap R^K$ contains an $L$-basis of~$A_i$.
Back to groups. If $\Char k$ does not divide $|G|$ and $M_i$ is a complete set of $kG$-simple
modules. Then $|G|\le\sum_i(\dim_kM_i)^2$ with equality iff $k$ is a splitting field.
\proclaim Definition. A set~$I\subset R$ is called nil if every element of~$I$ is nilpotent.
\proclaim Theorem. Any nil left ideal $J\subset R$ is nilpotent.
The sum of two nil ideals is also nil.
\proclaim Proof. Trivial.
The first part uses descending chain condition on left ideals and Nakayama lemma.
\proclaim Proposition. Let $\rad(R)$ (Wedderburn Radical) be the sum of all nil ideals of~$R$.
Then this is a nilpotent ideal that contains all nil left (or right) ideals J.
\proclaim Corollary. The ring $R/\rad(R)$ is semisimple.
The Wedderburn radical is the set of all elements that kill all simple modules.
\proclaim Proof. After replacing $R$ by $R/\rad(R)$ we may assume $\rad(R)=0$.
We want $_RR$ to be semisimple. Since it is artinian, it is sufficient to check
that every minimal left ideal~$E$ splits off.
Note that $E^2\ne0$, since otherwise it would be nil.
Hence there is an $a\in E$ such that $Ea\ne0$.
We must have $Ea=E$.
Write $a=ea$ where $e\in E$.
Consider left ideal $X=\{x\in E\mid xa=0\}$.
We have $X\subset E$ and $X\ne E$, hence $X=0$.
We have $a=ea=e^2a$, therefore $e=e^2$.
Then $E=Re$ splits.
Let $I=\rad(R)$. Want $IM=0$ for all simple~$M$.
Otherwise we have $IM=M$ and $M=I^NM=0$. Contradiction.
Let $\bar R=R/\rad(R)$. $\bar R$ is semisimple.
So simple $\bar R$-modules are certainly simple $R$-modules.
The direct sum of a complete set of simple modules over~$\bar R$ is faithful over~$\bar R$.
If $r$ kills all $R$-simples, then it kills all $\bar R$ simples, hence $\bar r=0$,
therefore $r\in\rad(R)$.
We can define general $\rad(R)$ as the set of all elements that kill all simple modules.
This is the Jacobson radical.
Nil radicals present an obstruction to semisimplicity.
Factorization by largest nil ideal yields a semisimple ring.
A ring and its factor ring by Wedderburn radical have the same simple modules.
Recall that the (Wedderburn) radical of a finite-dimensional algebra is the set of all
elements that kill all simple modules.
If we apply this definition for arbitrary ring, we obtain Jacobson radical.
\proclaim Theorem. Suppose that $k$ has characteristic~$p$ and $G$ is a finite $p$-group.
Let $M$ be a simple $kG$-module. Then $G$ acts trivially on~$M$. In particular, $\dim_kM=1$.
\proclaim Proof. Suppose that $|G|=p^n$. We use induction on~$n$.
Fix a central element of order~$p$.
Obviously, $D(c)-I$ is nilpotent. Hence $c$ acts trivially on its kernel.
Now view $M$ as simple $k[G/\langle c\rangle]$-module and induct.
\proclaim Corollary. Under same hypothesis, $\rad(kG)=I$, where $I$ is the augmentation ideal.
And $I^{|G|}=0$. Every proper left (right) ideal of~$kG$ is contained in~$I$.
So $kG$ is a noncommutative local ring.
\proclaim Proof. For every $g\in G$ the element $g-1$ acts trivially on every single $kG$-module.
Then $g-1\in\rad(kG)$. Therefore, $I\subset\rad(kG)$ and $I=\rad(kG)$. We know that this
ideal is nilpotent.
The composition series of $_{kG}kG$ has exactly $|G|$ factors.
We know that $I$ kills each composition factor.
Hence, $I^{|G|}=0$.
\proclaim Refinement. Suppose that $k$ has characteristic~$p$ and $G$ is a finite group
with a normal $p$-subgroup~$H$. Then $H$ acts trivially on any simple $kG$-modules.
\proclaim Proof. Let $M_0$ be the submodule consisting of all elements that are
invariant under action of center of~$G$.
It is a $kG$-submodule, hence it coincides with~$M$.
\proclaim Corollary. If $k$ has characteristic~$p$ and $G$ has a normal $p$-Sylow subgroup~$H$.
Then simple $kG$-modules are the same as simple $k(G/H)$-modules. We are back to non-modular
representations.
Moreover, $\rad(kG)=\sum_{h\in H}kG(h-1)$.
\proclaim Proof. First we verify that RHS is an ideal.
Then we verify that RHS is contained in $\rad(kG)$.
At last we observe that if we mod out RHS we obtain a semisimple ring $k[G/H]$.
Hence, $\rad(kG)$ is contained in RHS.
\section Chapter 2. Theory of Characters.
If $M$ is a left module over finite dimensional $k$-algebra,
such that $M$ is finite-dimensional over~$k$, then
its character is a function $\chi=(r\to\tr(m\to rm))$.
If $0\to M'\to M\to M''\to0$ is exact, then $\chi_M=\chi_{M'}+\chi_{M''}$.
\proclaim Theorem. Suppose that $k$ has characteristic zero and $R$ is finite-dimensional
$k$-algebra. Then isomorphism classes of semisimple $R$-modules are determined by their characters.
\proclaim Proof. Let $R/\rad(R)=\prod_iW_i$ be the Wedderburn decomposition.
Let $M_i$ be the corresponding $R$-simples.
Let $M=\oplus_il_iM_i$. We need to compute $l_i$ in terms of~$\chi_M$.
Fix $s\in R$ such that $s_j=[i=j]$. We have $\chi_M(s)=l_i\chi_{M_i}(s)=l_i\dim_kM_i$.
Since the characteristic is zero, $l_i$ is uniquely determined.
Back to groups.
If $k$ has characterstic zero, two representation of~$G$ are equivalent iff their
characters coincide.
\proclaim Proposition. In any characteristic if the order of~$g$ is equal to~$m$,
then $\chi_D(g)$ is equal to the sum of $m$th roots of unity in algebraic closure of~$k$.
In particular, if $k=\C$, then $\chi_D(g)$ is an algebraic integer in $\Q(\zeta_m)$.
\proclaim Definition. The kernel of character is the set of all group elements~$g$ such that
$\chi(g)=\chi(1)\in k$.
\proclaim Theorem. If $k$ has characteristic zero, then $\ker\chi=\ker(D)$.
If $\chi_i$ are all the characters of $kG$-simples, then the intersection of their
kernels is the trivial group.
\proclaim Proof. We can assume that $k$ is algebraically closed.
If $g\in\ker\chi$, then $n=\chi(1)=\sum_i\lambda_i$, where $\lambda_i$ are the eignevalues of~$D(g)$.
Clearly, $\lambda_i$. Moreover, $D(g)^{|G|}=1$, hence $D(g)$ is diagonalizable.
Hence $D(g)=1$.
\proclaim Definition. Let $\Irr(G)$ be the set of all characters of irredecuble representations.
For a character $\chi$ define its center as the set of all elements~$g$ such that $|\chi(g)|=\chi(1)$.
\proclaim Theorem. An element $g$ belongs to the center of a character iff $D(g)=\lambda I$
iff $g\in Z(G/\ker\chi)$.
\proclaim Proof. As before, conclude that $D(g)$ is diagonalizable and all of its eigenvalues
are equal to each other.
Conversely, if $g\in Z(G/\ker\chi)$ we conclude that $g$ belongs to the center.
\proclaim Notation. We always assume that $\Char k$ does not divide $|G|$.
We have $kG=\prod\Mat_{n_i}(D_i)$, $m_i=n_id_i$ etc.
Let $\chi_i$ be the $i$th irreducible character.
\proclaim Centrally primitive idempotents theorem. Let $e_i$ be the identities of the Wedderburn
components. These are centrally primitive idempotents in $kG$.
A central idempotent is called primitive if it is nonzero and we cannot represent
it as a sum of two nonzero central idempotents which have zero product (are orthogonal).
We have $e_i=n_i/|G|\sum_g\chi_i(g^{-1})g$.
In particular, $\Char k$ does not divide~$n_i$.
If, in addition, $k$ is a splitting field, then $C_g=|g|\sum_i\chi_i(g)e_i/n_i$.
\proclaim Proof. Suppose that $e_i=\sum_ha_{i,h}h$.
If $\chi_r$ is the regular character, then we have $\chi_r(e_ig^{-1})=a_{i,g}|G|$.
We have $a_{i,g}=|G|^{-1}\chi_r(e_ig^{-1})=|G|^{-1}n_i\chi_i(g^{-1})$.
To prove the second relation, note that $C_g=\sum b_{g,i}e_i$.
Applying $\chi_j$ on both sides, we have $|g|\chi_j(g)=b_{g,j}n_j$.
(Under splitting assumption we have $d_i=1$.)
Hence, $b_{g,j}=|g|\chi_j(g)/n_j$.
\proclaim Theorem. Frobenius integrality theorem.
Suppose $k$ has characteristic zero and is a splitting field.
Then $|g|\chi_i(g)/n_i$ is an algebraic integer.
Moreover, $\dim M_i=n_i$ divides $|G|$.
\proclaim Proof. The center of $kG$ is $\prod_ike_i$.
Now $C_g\in\Z G$ is a ring that is finitely generated as an abelian group.
Projecting upon the $ke_i$ we get images that are algebraic integers.
Finally, $e_i=n_i|G|^{-1}\sum_g\chi_i(g^{-1})g$.
We have $|G|/n_i\in\sum_gAC_g\subset\sum_gA\left(\sum_jAe_j\right)\sum_jAe_j$.
Hence $|G|/n_i$ is a rational integer.
\proclaim First orthogonality relation. (No splitting field assumption.)
For all $i$~and~$j$ we have $$|G|^{-1}\sum_{g\in G}\chi_i(g^{-1})\chi_j(hg)=[i=j]\chi_i(h)/n_i.$$
\proclaim Proof. Use the fact that $e_ie_j=[i=j]e_i$.
Recall that $e_i=n_i|G|^{-1}\sum_g\chi_i(g^{-1})g$.
Comparing coefficients of $h^{-1}$ we find out that
$n_i|G|^{-1}n_j|G|^{-1}\sum_g\chi_i(g^{-1})\chi_j(hg)=[i=j]n_i|G|^{-1}\chi_i(h)$.
\proclaim Corollary. If $k$ has characteristic zero (no splitting field assumption),
then a represetation is absolutely irreducible iff $\sum_g\chi(g^{-1})\chi(g)=|G|$.
\proclaim Proof. Say $\chi_D=\chi_i$ with $d_i=1$. Apply FOR with $i=j$.
To prove the converse, write $\chi=\sum_i p_i\chi_i$.
Then $|G|=\sum_g(\sum_ip_i\chi_i(g^{-1}))(\sum_jp_j\chi_j(g))
=\sum_{i,j}p_i^2d_i|G|$.
Since $\Char k=0$, we have $\sum_ip_i^2d_i=1$.
Hence $p_i=d_i=1$ for some~$i$ and $p_j=0$ for all other~$j$.
Now consider the set of all functions $\mu\colon G\to k$ that are constant on conjugacy classes.
Define a $k$-bilinear form $[\mu,\nu]=|G|^{-1}\sum_g\mu(g^{-1})\nu(g)\in k$.
\proclaim Corollary. Assume that $k$ is a splitting field.
Then $\chi_i$ form an orthonormal $k$-basis. Moreover, for any $\mu\in F_k(G)$ we have
$\mu=\sum_i[\mu,\chi_i]\chi_i$ (Fourier expansion).
We also have Plancherel formula: for all $\mu$~and~$\nu$ in $F_k(G)$ we have
$[\mu,\nu]=\sum_i[\mu,\chi_i][\nu,\chi_i]$.
Assuming characteristic zero we see that $\mu\in F_k(G)$ is of the form~$\chi_M$
for some $kG$-module~$M$ iff $[\mu,\chi_i]$ are all nonnegative integers.
Moreover, $M$ is irrdeucible iff $[\mu,\mu]=1$.
\proclaim Second orthogonality relation. Suppose that $k$ is a splitting field
and $g$~and~$h$ are two elements of~$G$. Then $\sum_i\chi_i(g)\chi_i(h^{-1})=[g\sim h]|C_G(g)|$.
\proclaim Proof. Use CPI and CPI for splitting field case.
\proclaim Applications to permutation characters.
\proclaim Burnside's lemma. The number of orbits of an action of a finite group~$G$ on a set~$E$
is equal to $[1,\pi]=|G|^{-1}\sum_g \pi(g)$, where $\pi(g)$ is the number of elements that
are fixed by the element~$g$.
\proclaim Proof. It is sufficient to prove the lemma for transitive case.
We have $|G|=n|G_i|$. Now $\sum_g\pi(g)=n|G_i|$.
\proclaim Theorem. Suppose $G$ acts transitively on~$E$ and let $t$ be the number of orbits of~$G_1$.
Then $t=[\pi,\pi]$.
\proclaim Proof. Expand $|G|\cdot[\pi,\pi]$.
\proclaim Lemma. Suppose $G$ is transitive on~$E$ and $n\ge2$.
Then $G$ is doubly transitive iff $t=2$ where $t$ is the number of $G_1$-orbits on~$E$.
Here $G_1$ is the stabilizer of an element.
\proclaim Theorem. Suppose $G$ is transitive on~$E$. Then $G$ is doubly transitive iff
$\bar V$ is an absolutely irreducible $kG$-module.
Here $\bar V$ is reduced $kG$-module corresponding to the factor of free $k$-module on~$E$
by all elements with zero sum.
\proclaim Proof. Let $\chi=\chi_{\bar V}=\pi-1$.
Then $[\chi,\chi]=[\pi-1,\pi-1]=[\pi,\pi]-2[\pi,1]+[1,1]=t-2m+1=t-1\in k$.
We know that $\bar V$ is absolutely irreducible iff $[\chi,\chi]=1$ iff $t=2$ iff $G$
is doubly transistive.
Now assume that $k$ is a splitting field. Let $g_i$ be a set of representatives of conjugacy
classes. The the character table $C_{i,j}=\chi_i(g_j)$.
Let $B_{i,j}=|g_i|\cdot|G|^{-1}\chi_j(g_i^{-1})$.
\proclaim Theorem. FOR holds iff $CB=1$.
SOR holds iff $BC=1$.
Hence, all the statements are equivalent to each other.
\proclaim Proof. Trivial substitution.
If $k=\C$ and $\mu=\sum_ia_i\chi_i$ for real~$a_i$, then $\overline{\mu(g)}=\mu(g^{-1})$.
On $\C$ one usually uses a different pairing: $\langle\mu,\nu\rangle=[\mu,\bar\nu]$.
This is a positive definite hermitean form.
Irreducible characters form an orthonormal basis as before.
We also have Fourier expansion and Plancherel formula.
\proclaim Corollary. For any class function $\chi\in F(G)$ define $\Q(\chi)=\sum_g\Q\chi(g)\subset\C$.
If $\chi$ is an irreducible character, then $\Q(\chi)$ is an algebraic number field.
This follows from char.pdf: for any irreducible character~$\chi$ we have
$\chi(g)\chi(h)=\chi(1)|G|^{-1}\sum_z\chi(gh^z)$.
\proclaim Theorem. Every $\chi\in\Irr(G)$ satisfies char.pdf: $\chi(g)\chi(h)=\chi(1)|G|^{-1}
\sum_z\chi(gh^z)$, where $h^z=z^{-1}hz$.
\proclaim Corollary. If $\chi\in\Irr(G)$, then $\Q(\chi)$ is an abeliean field extension of~$\Q$.
\proclaim Proof of Corollary. $\Q(\chi)$ is a finite-dimensional $\Q$-domain, hence a field extension
of~$\Q$. Observe that $\chi(g)\in\Q(\zeta)$ where $\zeta=\exp(2\pi i|G|^{-1})$.
We see that $\Q(\chi)$ is Galois over~$\Q$ with Galois group $\Gal(\Q(\chi)/\Q)=G/H$ which is
an abelian group.
\proclaim Proof of Theorem. For $g\in G$ define $\alpha_{i,j,g}=\#\{(g',g'')\in G\times G\mid g=g'g''\land g'\sim g_i\land g''\sim g_j\}$.
Note that $g\mapsto\alpha_{i,j,g}$ is a class-function.
Let's write $\alpha_{i,j,p}=\alpha_{i,j,g}$ where $g$ belongs to $p$th conjugacy class.
Then $C_iC_j=\sum_p\alpha_{i,j,p}C_p$.
Apply $\pi_l$ to $C_iC_j$.
We get $|g_i|\cdot|g_j|\chi_l(1)^{-1}\chi(g_i)\chi(g_j)=\sum_p\alpha_{i,j,p}|g_p|\chi(g_p)$.
\proclaim Converse Theorem. An aribtrary function $\mu\colon G\to\C$ is a scalar
multiple of irreducible character iff $\mu$ satisfies char.pdf.
\proclaim Sketch of Proof. Last part: write $\mu=z\chi$ where $\chi\in\Irr(G)$.
We have $\mu(1)=z\chi(1)\in\R^+$. Hence $z\in\R^+$. Also $1=\langle\mu,\mu\rangle=z^2\langle\chi,\chi\rangle=z^2$.
Hence $z=1$ and $\mu=\chi$.
First part: suppose that char.pdf holds. Suppose $\mu(1)=0$. Then $\mu=0\cdot1_G$.
Assume $\mu(1)\ne0$. In char.pdf set $g=1$. We have $\mu(h)=|G|^{-1}\sum_z\chi(h^z)$.
Hence $\mu$ is a class function. Define $\pi\colon Z(\C G)\to\C$ by $\pi(C_i)=|g_i|\mu(g_i)\mu(1)^{-1}\in\C$.
Check that $\pi$ is a $\C$-algebra homomorphism. It is enough to check
that $\pi(C_i)\pi(C_j)=\sum\alpha_{i,j,p}\pi(C_p)$ via char.pdf.
After this, $\pi=\pi_l$ for some $\mu$. Then you check $\mu=z\chi_l$.
\section New Representations from Old.
(1)~Twist a group representation by group automorphism. Only outer automorphisms yield nontrivial twistings.
(2)~Twist a group representation by field automorphism.
(3)~Twist a group character by field automorphism of character field.
In the last case we obtain a group character because char.pdf stays true under field automorphism.
To connect (2)~and~(3) note that we can extend (non-uniquely) a field automorphism of a character
field and obtain the twisted representation.
(4)~If $V$ is a $kG$-module, then $V^*$ is also a $kG$-module: $(g\lambda)(v)=\lambda(g^{-1}v)$.
Obviously, $\chi_{D'}(g)=\chi_D(g^{-1})$ for all~$g$.
Note that the group inverse sends conjugacy classes to conjugacy classes.
We say that a conjugacy class is real if it is invariant under group inverse.
A character over complex numbers is real-valued if all of its values are real.
\proclaim Burnside Theorem 1. The number of real conjugacy classes is equal
to the number of real-valued characters.
\proclaim Proof (Brauer). By permuting pairs of complex-conjugated characters we get a matrix
$PC$. Similarly, by permuting pairs of conjugacy classes which are mutually inverse
we get a matrix~$CQ$.
Now $C^{-1}PC=Q$ and $\tr(P)=\tr(Q)$.
\proclaim Corollary, $|G|$ is even iff the is a real-valued irreducible character~$\chi\ne1$.
\proclaim Proof. From Burnside's theorem we obtain that the existence of~$\chi$ is equivalent
to existence of self-inverse conjugacy class.
\proclaim Proposition. Suppose that the columns of character table are permuted arbitrarily.
Then we can compute which column corresponds to the identity class. Moreover we can determine the
sizes of conjugacy classes.
\proclaim Proof. Suppose that $\chi\in\Irr(G)$. Note that $\ker\chi=\{g\in G\mid\chi(g)=\chi(1)\}$.
We will also prove that $\ker\chi=\{g\in G\mid0<\chi(g)\ge|\chi(h)|\}$.
One of the inclusions is clear. The other inclusion is also trivial.
Now recall that the intersection of the kernels of all characters consists of trivial element.
Hence, the first column is uniquely determined by the character table.
Now we can find $|g_j|$ by second orthogonality relation.
\proclaim Theorem. The character table determines the position of the elements of the group
commutant.
\proclaim Proof. Recall that the commutant is equal to the intersection of all one-dimensional
characters.
\proclaim Theorem. The character table determines all normal subgroups.
\proclaim Proof. Denote by $N_i=\ker\chi_i$ a set of normal subgroups.
Obviously, their finite intersections constitute the set of all normal subgroups.
\proclaim Theorem. $G$ is simple iff all kernels are trivial except for the kernel corresponding to the first row.
\proclaim Proof. Trivial consequence of the previous theorem.
\proclaim Theorem. Given a normal subgroup of given group~$G$, the character table of~$G$
determines the character table of factor group (although not its isomorphism class).
\proclaim Proof. Take all irreducible characters whose kernel contains given normal subgroup.
These characters correspond to all irreducible characters of factor group.
Now remove all duplicate columns.
\proclaim Theorem. The character table determines the position of the center.
It also determines whether the group is abelian and its isomorphism class is uniquely determined.
\proclaim Proof. The center is the set of all elements whose conjugacy class size is~1.
It is also the intersection of the centers of all characters.
\proclaim Theorem. The character table determines the position of upper central series.
Hence it determines whether the group is nilpotent.
\proclaim Proof. Trivial consequence of definitions and previous results.
\proclaim Theorem. The character tables determines whether the group is solvable.
\proclaim Proof. A group is solvable iff there is a normal series with $p$-groups as factors.
\proclaim Proposition. If $H$ is a normal subgroup of~$G$, then $|C_{\bar G}(\bar g)|\le|C_G(g)|$.
\proclaim Second Burnside theorem. If $|G|$ is odd, then it is congruent
to the number of conjugacy classes modulo~16.
\proclaim Proof. The only real irreducible character is the trivial character.
Also, the dimensions of irreducible characters are odd, since they divide
the order of group.
\proclaim Corollary. If $|G|\le19$ is odd, then it is odd.
\proclaim Theorem. If $|G|$ is odd and every prime that divides~$|G|$ is
congruent to~1 modulo~4, then $|G|$ and the number of conjugacy classes
are congruent modulo~32.
\section Chapter 3. Tensor Products and Invariants.
Multiplicative structure on character ring provides a good tool for computation.
We can construct new irreducible representations from existing ones. For example,
we can now determine the character table of $A_5$ given only its icosahedral representation.
Another kind of product (outer product): Let $V$ be a $kG_1$-module and $W$ be a $kG_2$-module.
Then $V\#W=V\otimes_kW$ is a $kG$-module, where $G=G_1\times G_2$.
Here $(g_1,g_2)(v\otimes w)=g_1v\otimes g_2w$.
This construction is a special case of tensor product if we regard $V$~and~$W$ as $kG$-modules.
\proclaim Proposition. Suppose that $V$~and~$V'$ are irreducible $kG_1$-modules
and $W$~and~$W'$ are irreducible $kG_2$-modules. Then $V\#W$ is isomorphic to $V'\#W'$ iff $V$ is
isomorphic to $V'$ and $W$ is isomorphic to $W'$.
If $V$ and $W$ are absolutely irreducible, then $V\#W$ is also absolutely irreducible.
\proclaim Proof. The first part is trivial if we regard $V\#W$ as $kG_1$-module and decompose it into
irreducible parts. This implies that $V$ is isomorphic to~$V'$. The same argument goes for~$W'$.
The second part is proven as follows. Recall that $kG$ is isomorphic to $kG_1\otimes kG_2$.
By Burnside's theorem the maps $D_1\colon kG_1\to\End_k(V)$ and $D_2\colon kG_2\to\End_k(W)$
are epimorphisms. Hence $\End(V\#W)$ is isomorphic to $\End(V)\otimes\End(W)$.
Therefore the map $kG\to\End(V\#W)$ is epimorphic, hence $V\#W$ is absolutely irreducible.
\proclaim Theorem. Suppose that $k$ is the splitting field for $G_1$ and $G_2$.
Suppose that $\Char k$ does not divide $|G|$.
Tensor products of all pairs of irreducible representations of $G_1$~and~$G_2$ form
complete set of irreducible representations of~$G_1\times G_2$.
\proclaim Proof. Obviously all of these representations are non-isomorphic and absolutely irreducible.
The number of conjugacy classes of direct product of groups is equal to the product
of the corresponding numbers of conjugacy classes.
Note that the character table of direct product of groups is equal to the tensor product
of corresponding character tables.
In particular, if some representations have the same character tables, then their
tensor products with anything also have the same character tables.
Let $V$ be a simple $\C G$-module and let $\chi=\chi_V$.
\proclaim Frobenius theorem. $\chi(1)$ divides $|G|$.
\proclaim Schur's theorem. $\chi(1)$ divides $[G:Z(G)]$.
\proclaim Generalized Schur's theorem. $\chi(1)$ divides $[G:Z(\chi)]$.
\proclaim Proof. Recall that $H=Z(\chi)=\{h\in G\mid|\chi(h)|=\chi(1)\}$ is a normal subgroup of~$G$.
Note that $V^n=\#^nV$ is an irreducible $\C G^n$-module.
Also $H$ acts on $V$ as complex numbers: $hv=\lambda(h)v$.
Now we see that $K_n=\{(h_1,\ldots,h_n)\in H^n\mid\lambda(\prod_ih_i)=1\}$ is a normal subgroup
of $G^n$ that acts trivially on~$V^n$.
Hence $V^n$ is a simple $G^n/K_n$-module.
Note that $|H|^{n-1}$ divides $K_n$ because projection of $K_n$ onto first $n-1$ arguments is surjective.
Now $|G^n/K_n|=|H|\cdot[G:H]^n/x$.
By Frobenius this is divisible by $\chi(1)^n$.
Hence $\chi(1)$ divides $[G:H]$.
\proclaim Burnside-Brauer theorem. Assume that $\Char k=0$ and $D$ is an arbitrary representation
of~$G$. Suppose that the character of $D$ is faithful and takes $m$ distrinct values.
Then any irreducible character is contained in $\phi^i$ for $0\le i0$, then $|P|=p^r$.
\proclaim Proof. $G$ is nonabelian. Take $g\in P$ such that $g\ne1$.
Then $\chi$ is prime to~$g$. Hence $\chi(g)=0$. Therefore, $|g|$ is a $p$-prime number.
Now $\langle\chi_P,1_P\rangle=|P|^{-1}\chi(1)$.
Hence $|P|=\chi(1)=p^r$.
\proclaim Proposition. If $G$ is a simple group, $\chi$ is an irreducible character,
$\chi(1)=p$ where $p$ is a prime number, then the $p$-part of $|G|$ is $p$ and $p\ne2$.
\proclaim Proof. It suffices to show that $p$-Sylow subgroup $P$ is abelian.
Since $p$ divides $|G|$, we have $P\ne\{1\}$. Recall that $G$ acts faithfully on~$V$
and $Z(\chi)=Z(G)=\{1\}$.
If $V_P$ is a simple $\C P$-module, then
$1\ne Z(P)\subset Z(\chi_P)=\{g\in P\mid|\chi_P(g)|=p\}\subset Z(\chi)=\{1\}$.
Hence $V_P$ is not simple, hence any 2 elements of $P$ thus commute.
\proclaim Classification of finite simple groups.
Series of simples groups: cyclic groups of prime order,
$A_n$ for $n\ge5$, linear groups (Jordan, Dickson) and other groups of Lie type by Chevalley,
sporadic simple groups by Mathiue and others ending at Monster, 26 in total.
Burnside was a pioneer in this area. Brauer was a visionary.
His idea was to study the centralizer of an involution (element of order 2).
Feit and Thompson (young hotshots) proved that odd order group are solvable.
Gorenstein was the field marshall (20 year war).
Relationship between simple groups and the prime 2:
$\chi(1)\ne2$, the number of distinct primes divisors of $|G|$ is not 2,
and 2 divides $|G|$.
Brauer'sprogram of classifying finite simple groups: prove theorems like
``if $G$ is a finite simple group that has centralizer of involution isomorphic to a given
group~$T$ then $G$ is isomorphic to one of the finite number of given groups''.
We will illustrate this philosophy in the simplest case:
\proclaim Theorem. Suppose we have an involution $u$ whose centralizer has degree~2.
Then $[G:[G,G]]=2$. In particular, if $G$ is simple, then it has order~2.
\proclaim Proof. We know that $\chi_i(u)$ is the sum of eigenvalues. Every eigenvalue for~$u$ is 1~or~$-1$.
On the other hand we have SOR $\sum_i\chi_i(u)\chi_i(u^{-1})=|C_G(u)|=2$.
Hence $2=\sum_i\chi_i(u)^2$. Say, $\chi_2(u)=\pm1=\epsilon$ and all other $\chi_i(u)=0$.
We have another SOR: $0=1\cdot1+\epsilon\chi_2(1)$. Hence $\chi_2(1)=1$ and $\epsilon=-1$.
A linear character cannot have~0 as a value.
Hence we have 2 linear characters and $[G:[G,G]]=2$.
\proclaim Important tool of Frobenius-Schur indicator.
\proclaim Theorem.
For arbitrary representation~$V$ of finite group~$G$
with character~$\chi$ we have
$\chi_{S^2V}(g)=(\chi(g)^2+\chi(g^2))/2$ and $\chi_{\Lambda^2V}(g)=(\chi(g)^2-\chi(g^2))/2$.
\proclaim Definition. (Frobenius-Schur Indicator.) Let $k$ be
an algebraically closed field of characteristic~0. Then $s(\chi)=|G|^{-1}\sum_g\chi(g^2)
=|G|^{-1}\sum_g(\chi_{S^2V}-\chi_{\Lambda^2V}(g))=\langle\chi_S,1\rangle-\langle\chi_A,1\rangle$
is an integer number.
Note that $\langle\chi_S,1\rangle+\langle\chi_A,1\rangle=\langle\chi^2,1\rangle
=|G|^{-1}\sum_g\chi(g)\chi(g)=\langle\chi,\bar\chi\rangle=[\hbox{$\chi$ is real}]$.
\proclaim Corollary. For $\chi\in\Irr(G)$ there are exactly 3 possibilities:
\halign{
&$#$\hfil\cr
\langle\chi_S,1\rangle&\langle\chi_A,1\rangle&s(\chi)\cr
1&0&1\cr
0&1&-1\cr
0&0&0\cr
}
Hence we have 3 possible types of irreducible characters.
We can call them type 1, type 2, and type 3 charcters.
An example of dihedral group of order~8 and quaternion group shows
that $s(\chi)$ cannot be determined from the character table.
Abelian group does not have type 2 characters.
A nontrivial character of odd group is always indefinite.
\proclaim Frobenius-Schur theorem. An irreducible character has type 1 iff $V$ is defined over reals.
\proclaim The number of square root of a group element is a class function, which is a virtual
character: $\theta=\sum_\chi s(\chi)\chi$.
The number of involutions in~$G$ is equal to $t=\sum_{\chi\ne1}s(\chi)\chi(1)$.
Moreover, $t^2\le(s-1)(n-1)$, where $n=|G|$ and $s$ is the number of conjugacy classes.
\proclaim Proof. First note that $\langle\theta,\chi\rangle=|G|^{-1}\sum_g\theta(g)\chi(g^{-1})=|G|^{-1}\sum_h\chi(h^{-2})=s(\chi)$.
Next note that $1+t=\theta(1)=\sum_\chi s(\chi)\chi(1)=1+\sum_{1\ne\chi\in\Irr(G)}s(\chi)\chi(1)$.
And now $t^2\le(\sum_{\chi\ne1}\chi(1))^2=(\sum_{\chi\ne1}1\cdot\chi(1))^2\le(s-1)\sum_{\chi\ne1}\chi(1)
=(s-1)(n-1)$.
\proclaim Corollary. If the group is even, then there is a non-trivial conjugacy class of size
not exceeding $((n-1)/t)^2$.
\proclaim Brauer-Fowler theorem. Let $m$ be an integer number and $G$ be a simple group.
Let $u$ be an involution such that $|C_G(u)|\le m$.
Then $|G|<(m^2)!$. In particular there exists only finitely many such groups~$G$.
This result motivated Brauer's program.
\proclaim Involution count formula. The number of involutions equals $\sum_{\chi\ne1}s(\chi)\chi(1)$.
\proclaim Proof of Brauer-Fowler theorem. Let $n=|G|$. Then $t\ge|u|=|G|\cdot|C_G(u)|^{-1}\ge n/m$.
Existence of small class theorem implies that there is $g\ne1$ such that $|g|<(n/t)^2\le m^2$.
We have a small conjugacy class. The group is simple, hence it must act faithfully on the cosets
of this class, therefore we can embed it into symmetric group of required size.
More precisely, if $H=G$, then $1\ne g\in Z(G)$, therefore $G=Z(G)$ and $|G|=2$.
If $H\ne G$, then $r=[G:H]>1$.
And $G$ maps to the symmetric group of the set $G/H$, which has more than 1 element.
This action is faithful.
Brauer's vision was to study finite simple groups by looking at and controlling the centralizer of an involution.
\section Chapter 5. Induced representations.
Two ways to get induced representations: restriction and tensor product (over noncommutative rings).
If $U$ is a $kG$-module and $H\subset G$, then $U_H$ is $kH$-module obtained by restriction.
If $V$ is a $kH$-modules, then $V^G$ is a $kG$-modules obtained by tensor product.
Now $(kG)_{kH}$ is free of finite rank $[G:H]$.
In particular, $\dim(V^G)=[G:H]\dim V$.
Suppose $V$ is 1-dimensional (over~$k$) $kH$-module. Then $V^G$ is called monomial $kG$-module.
If we use matricial representation, we see that $D^G$ is represented by block monomial (generalized
permutation) matrices.
We can easily see that $\chi^G(g)=\sum_i\dot\chi(g_i^{-1}gg_i)$. Here $\dot\chi$ is $\chi$
extended by zeros. Hence $\chi^G(g)=0$ whenever $g$ does not belong to any conjugate of~$H$.
In particular, if $H$ is a normal subgroup of~$G$, then $\chi^G(G\setminus H)=0$.
Moreover, if $H\subset Z(G)$, then $\chi^G=[G:H]\dot\chi$.
Note that in general $\dot\chi$ is not a class function on $G$. We can only say that $\dot\chi$
does not depend on $H$-conjugacy class.
\proclaim Corollary. If $\Char k$ does not divide $|H|$, then $\chi^G(g)=|H|^{-1}\sum_{t\in G}
\dot\chi(t^{-1}gt)$.
\proclaim Philosophy. Many irreducible representations of~$G$ are monomial.
\proclaim Lemma. If $f\colon S\to R$ is a ring homomorphism. If $V$ is an $S$-module, then
$V^R=R\otimes_SV$ is an $R$-module. There is a natural abelian group isomorphism
between $\Hom_R(V^R,U)=\Hom_S(V,U_S)$.
\proclaim Proof. Obvious.
\proclaim Frobenius reciprocity. Suppose that $\Char k$ does not divide $|G|$. (Semisimple situation.)
If $H\subset G$, $U$ is a simple $kG$-module, $V$ is a simple $kH$-module,
$m$ is the number of copies of $V$ in $U_H$, $n$ is the number of copies of $U$ in $V^G$.
Then $m>0$ iff $n>0$. Also $m=n$ iff $\dim_k\End_{kG}U=\dim_k\End_{kH}V$.
The last assumption holds when $U$ and $V$ are absolutely irreducible (the dimension is 1
of endomorphism ring is 1 for such modules).
\proclaim Proof. The first statement follows from the previous lemma. The second one follows from
decomposition into irreducible modules. We have $m\End_{kH}V=n\End_{kG}U$.
\proclaim Corollary. If $\Char k$ does not divide $|G|$, $H\subset G$, $k$ is a splitting field
for $G$ and $H$, and $U_i$ and $V_j$ are all simple $kG$ and $kH$ modulesw, then
$(U_i)_H=\oplus_ja_{i,j}V_j$ iff $V_j^G=\oplus_ia_{i,j}U_i$.
\proclaim Theorem. In semisimple case without splittig field assumption we have $i(W,U)=[\chi_W,\chi_U]\in k$,
where $W$ and $U$ are finite-dimensional $kG$-modules. Here $i(W,U)=\dim\Hom_{kG}(W,U)$ is the intertwining
number. Moreover, if $U$ is a $kG$-module and $V$ is a $kH$-module, then $[\chi_U,\chi_V^G]=[(\chi_U)_H,\chi_V]$.
\proclaim Proof. Using additivity reduce everything to simple modules. Apply first orthogonality relation.
\proclaim Theorem. (No splitting field assumption.) If $U$ is a $kG$-module and $V$ is a $kH$-module,
then $V^G\otimes_kU=(V\otimes(U_H))^G$ as $kG$-modules.
Taking characters we obtain $(\nu^G)\mu=(\nu\mu_H)^G$.
\proclaim A couple of quick applications. Suppose that $\Char k=0$ and $G$ acts transitively on~$E$.
Identify $E$ with $G/H$ where $H$ is the stabilizer of a group element.
Then $kE$ is the induced representation of the trivial representation of~$H$.
Let $\pi$ be the fixed point counter: $\pi=1_H^G$. Now $[\pi,1]_G=[1_H^G,1]_G=[1,1]_H=1$ (Burnside lemma),
$[\pi,\pi]=[1_H^G,\pi]_G=[1,\pi_H]_H$ is the number of $H$-orbits on $E$. This is another old result.
If $H$ is a subgroup of abelian group~$G$. Then any linear character~$\nu$ of $H$ extends to a character of~$G$.
Proof: $\C^*$ is a divisible, hence injective $\Z$-module.
Character proof: $\nu^G=\oplus\mu_i$, hence $\nu=(\mu_i)_H$.
\proclaim Restriction and induction for class functions.
We assume that $\Char k$ does not divide $|G|$.
Let $F_k(G)$ be the $k$-algebra of class functions on~$G$.
Restriction: $F_k(G)\to F_k(H)$ is a ring homomorphism.
Induction: if $\nu\in F_k(H)$, then $\nu^G\colon G\to k$ is defined by $\nu^G(g)=|H|^{-1}\sum_t\dot\nu(t^{-1}gt)
=|H|^{-1}\sum_t\dot\nu(g^t)$. Here $\dot\nu$ is $\nu$ extended by zero.
Note that if $g\notin\cup H^t$ then $\nu^G(g)=0$. Here $\dot\nu=0$ is an extension of $\nu$.
If $\nu$ is a character, then $\nu^G$ is the induced character.
\proclaim Proposition. If $\nu\in F_k(H)$, then $\nu^G\in F_k(G)$. $(\nu^E)^G=\nu^G$.
$[\nu^G,\mu]_G=[\nu,\mu_H]_H$. $(\nu\mu_H)^G=\nu^G\mu$.
\proclaim Proof. Trivial.
\proclaim Definition. A group $G$ is called monomial group if every irreducible complex representation of $G$
is monomial.
\proclaim Examples. $Q_8$, $D_n$, $A_4$, $S_4$, $S_3$, special group of order 21.
\proclaim Non-monomial groups. $\rm BT_{24}$ has 2-dimensional irreducible character,
which is not monomial because the group does not have index-2 subgroup.
$A_5$ is not monomial because three of its characters that have order 3~and~4 are
not monomial because $A_4$ has no subgroup of index 3~and~4. Similarly, $S_5$ is not monomial.
\proclaim Two facts about monomial groups. If $|G|<24$, then it is monomial.
Nilpotent groups are monomial. Monomial groups are solvable.
A group is nilpotent iff it is a direct product of Sylow $p$-groups.
Monomial groups are closed under direct products. $p$-groups are monomial.
\proclaim Definition. A Hall subgroup is a subgroup $H$ such that $|H|$ and $[G:H]$ are
relatively prime. This is a generalization of Sylow subgroup.
\proclaim Third application. If $H$ is a Hall subgroup and $h\in H\cap Z(G)$. Then
$h\in G'$ iff $h\in H'$.
\proclaim Proof. Assume that $h\in G'$ and $h\notin H'$. Fix a 1-dimensional $\C H$-module~$V$
on which $h$ acts nontrivially. Fix a simple $\C G$-module $U\subset V^G$ whose
dimension is relatively prime to~$p$. This is possible only in Sylow case.
From Frobenius reciprocity it follows that $V$ is a simple submodule of $U_H$.
Now $h$ acts on $U$ by scalar multiplication by some $\lambda\in\C^*$.
Also the dimension of $U$ and $|H|$ are relatively prime.
If $h\in G'$ then $\det(D(h))=\lambda^{\dim U}=1$. On the other hand $\lambda^{|H|}=1$.
\proclaim Goal for the rest of the course. To prove the following
application of induced representations: If $G$ acts transitively on
set~$E$ such that any non-indentity element has at most one fixed point,
then for any two fixed-point free elements their product is also
fixed-point free unless it is equal to~1.
\proclaim Mackey theorems. Suppose $H$~and~$K$ are subgroups
of~$G$, $V$ is a $kH$-module. (1)~What can we say about $(V^G)_K$?
(2)~When can we say $V^G$ is simple.
If $G=\cup_ig_iH$, then $V^G=\oplus_ig_i\otimes V$.
We have $V=1\otimes V\subset V^G$.
In fact $gV=g(1\otimes V)=g\otimes V$.
Note: $G$ permutes the $gV$'s transitively. (If $g\in g_iH$, then $gV=g_iV$.)
Also $H$ is the isotropy subgroup of~$V$.
\proclaim Recognition criterion for an induced module. ($H$ is not
given a priori.) Suppose $U$ is a $kG$-module such that $U=\oplus_iV_i$,
where $V_i$ are $k$-vector subspaces. Suppose $G$ acts on $U$
in such a way that $U_i$ are permuted transitively.
Let $H$ be an isotropy subgroup of~$V$.
Then $V$ is a $kH$-module and $U=V^G$.
\proclaim Proof. It is sufficient to prove that $V^G=kG\otimes_{kH}V\to U$
is surjective and two modules have the same dimension.
\proclaim Theorem. Suppose $H$~and~$K$ are subgroups
of~$G$, $V$ is a $kH$-module. Given $K$ and $H$ write the double-coset
decomposition $G=\cup_{s\in S}KsH$.
For any $s\in G$ define $K_s=sHs^{-1}\cap K$.
Then $sV\subset V^G$ is a $kK_s$-submodule.
We have $(V^G)_K=\oplus_{s\in S}(sV)^K$.
\proclaim Proof. Define $V(s)=\sum_{g\in KsH}gV\subset V^G$.
Note that $V(s)$ is a $kK$-submodule.
Now we check that $V(s)=(sV)^K$ as $kK$-modules.
To see this, apply induced module criterion: $K$ acts
transitively on $\{gV\mid g\in KsH\}$.
Now the isotropy subgroup of $sV$ is: $xsV=sV$ iff $x\in sHs^{-1}\cap K=K_s$.
\proclaim Theorem. Mackey's irreducibility criterion.
We assume that $\Char k=0$. $V^G$ is absolutely irreducible iff
$V$ is absolutely irreducible and for any $s$ in $S\setminus\{1\}$
two $H_s$-modules $sV$ and $V_{H_s}$ have no common composition factors.
\proclaim Proof. We write $[U,W]$ for $[\chi_U,\chi_W]$.
$[V^G,V^G]=[V,(V^G)_H]_H=[V,\oplus_{s\in S}(sV)^H]_H=\sum_{s\in S}[V,(sV)^H]_H
=\sum_{s\in S}[V_{H_s},sV]_{H_s}=[V,V]_H+\sum_{s\in S\setminus\{1\}}[V_{H_s},sV]_{H_s}\in k$. Now $V^G$ is absolutely irreducible iff $[V^G,V^G]=1$
iff $[V,V]_H=1$ and for all $s\in S\setminus\{1\}$ we have $[V_{H_s},sV]_{H_s}=0$.
\proclaim Corollary. If $\Char k=0$ and $H$ is a normal subgroup of~$G$, then
Write $G=\cup_ig_iH$. Let $V$ be an absolutely irreducible $kH$-module.
Then $V^G$ is absolutely irreducible iff for any $i\ge2$ we have $V\ne g_iV$ as $kH$-modules.
\proclaim Corollary. If $\Char k=0$. Let $\nu\colon H\to k^*$ be a linear character.
Then the monomial character $\nu^G$ is absolutely irreducible iff for any $s\in S$ such that
$s\ne1$ there is an $h\in sHs^{-1}\cap H$ such that $\ne(h)\ne\nu(s^{-1}hs)=\nu(h^s)$, i.e.,
iff $\nu(h^{-1}s^{-1}hs)\ne1$.
\section Chapter 6. Frobenius groups.
If a finite group~$G$ acts on a finite nonempty set~$E$. For $e\in E$ we denote by~$G_e$
the isotropy subgroup of~$e$. We can easily see that $G_{ge}=g^{-1}G_eg$.
If $G$ is transitive on~$E$, then all isotropy subgroups are conjugate. Recall
that in transitive case $E=G/G_e$ as $G$-sets.
\proclaim Definition. We say that $G$-action on~$E$ is semiregular or regular
iff for any $e\in E$ we have $G_e=\{1\}$ (also called free) or the previous condition holds
and it is transitive.
\proclaim Definition. If $E$ is a transitive $G$-set, then let
$K=\{1\}\cup\{g\in G\mid\pi(g)=0\}=G\setminus\cup_eG_e$,
where $\pi$ is fixed point counter. $K$ is called the Frobenius kernel of the action.
It is closed under inversion and conjugation. If $K$ is closed under multiplication,
then it is a normal subgroup.
\proclaim Jordan's Inequality. If $G$ acts transitively on~$E$, then $|K|\ge|E|$.
\proclaim Proof. Burnside: $|G|=\sum_g\pi(g)=\sum_{g\in K}\pi(g)+\sum_{g\in G\setminus K}\pi(g)
\ge|E|+|G|-|K|$.
\proclaim Jordan's theorem. If $H$ is a subgroup of~$G$, then $|G\setminus\cup H^g|\ge[G:H]-1$.
\proclaim Proposition. Consider statements: (1)~$\pi(g)\le1$ for any $g\in G\setminus\{1\}$.
(2)~$|K|=|E|$. (3)~$K$ is a subgroup of~$G$ (hence $K$ is a normal subgroup of~$G$).
We have (1) iff (2). If (3) holds, then there exists an $H$-set isomorphism $K=E$
where $H$ acts on~$K$ by conjugation. In particular, (3) implies (1)~and~(2).
\proclaim Proof. (1) iff (2) is clear from proof of Jordan inequality.
Assume (3) holds. Define $\theta\colon K\to E$ such that $\theta(g)=ge$ where $g\in K$.
Now we check that this is an $H$-set morphism. If $g\in K$ and $h\in H$, then
$\theta({}^hg)=\theta(hgh^{-1})=(hgh^{-1})(e)=h(ge)=h\theta(g)$. Now we prove that
$\theta$ is injective.
\proclaim Definition. (Frobenius.) If (1) holds, we say that $G$-action on~$E$ is Frobenius
provided that $1<|E|<|G|$. In this case $G$ is a Frobenius group.
\proclaim Big Frobenius theorem. Let $E$ be a Frobenius $G$-set. Then $K$ is a subgroup.
This also means that (1),~(2),~and~(3) are equivalent.
\proclaim Corollary. Let $n=|E|$, where $E$ is a Frobenius $G$-set.
Then $K$ acts regularly on~$E$, $G$ is a semidirect product of $K$~and~$H$,
$H$ acts semiregularly on $E\setminus\{e\}$ and $K^*=K\setminus\{1\}$,
$|H|$ divides $n-1$, $K$~and~$H$ are Hall subgroups of~$G$,
$K=\{x\in G\mid x^n=1\}$.
\proclaim Corollary. $p$-groups, abelian groups, and simple groups cannot be Frobenius.
$K$ is called the Frobenius kernel, $H=G_e$ is called Frobenius complement, and we
have $|G|=|K|\cdot|H|$. Big Frobenius states that $K$ is a subgroup of~$G$ (hence a normal
subgroup).
Given a subgroup $H$ of $G$, when is $G/H$ a Frobenius $G$-set?
\proclaim Definition. A nontrivial subgroup $H$ of $G$ is said to have trivial
intersection property if for any $g\in G\setminus H$ we have $H\cap H^g=\{1\}$.
\proclaim Theorem. If $H$ is a subgroup of $G$, then $G$ acts Frobeniusly on $G/H$
iff $H$ has trivial intersection property.
\proclaim Proof. Assume $G/H$ is Frobenius. Then $1\ne H\ne G$. Consider $g\notin H$.
Now $H^g\cap H$ fixes $e$ and $g^{-1}e$, which are different points, hence $H^g\cap H=\{1\}$.
Now assume that $H$ has trivial intersection property. Suppose $g\ne1$ fixes cosets
$xH$ and $yH$. We have $xH=gxH$, hence $g\in H^{x^{-1}}$. Also $g\in H^{y^{-1}}$.
Honce $H^{x^{-1}}\cap H^{y^{-1}}\ne\{1\}$. Conjugate this by $y$.
We have $\{1\}\ne H^{x^{-1}y}\cap H$. By trivial intersection property $x^{-1}y\in H$,
hence $xH=yH$.
\proclaim Example. If $G$ is nonabelian group of order $pq$, where $p