14 250527

weak form of Clifford theorem

Let $k$ be a field, and let $U$ be a simple $kG$-module. For a normal subgroup $N⊲G$, ${\mathrm{Res}}_N^G(U)$ is semisimple as a $kN$-module.

\begin{proof} Let $V$ be a simple $kN$-submodule of ${\mathrm{Res}}_N^G(U)$. For every $g\in G$, $gV$ is a $kN$-module, because

$$n(gv)=g(n^g)v=g(n^{\prime }v).$$

Also note that $gV$ is also simple. Consider ${\sum }_{g\in V}gV$ is a $G$-invariant subspace of $U$. Since $U$ is simple, we know $U={\sum }_{g\in G}gV$. So ${\mathrm{Res}}_N^G(U)$ is a semisimple $kN$-module. \end{proof}

Recall that ${}^{g}V$ is a $k({}^{g}N)$-module, because ${}^{g}V\to gV,v\mapsto gv$ is a $kN$-isomorphism. For any $n\in N$, the action on ${}^{g}V$ is $nv=n^{g}v$, and the action on $gV$ is $n(gv)=g(n^{g}v)$.

Clifford theorem

Let $k$ be a field, and let $U$ be a simple $kG$-module. For a normal subgroup $N⊲G$, write ${\mathrm{Res}}_N^{G}U=S_1^{a_1}\oplus \cdots \oplus S_r^{a_r}$ with $S_i$ simple $kN$-modules, occurring with multiplicities $a_1,\cdots ,a_r$, respectively.

• $G$ permutes $\{S_1^{a_1},\cdots ,S_r^{a_r}\}$ transitively;
• $\dim S_1=\cdots =\dim S_r$, $a_1=\cdots =a_r$;
• if $H={\mathrm{Stab}}_G(S_1^{a_1})$, then $U={\mathrm{Ind}}_H^G(S_1^{a_1})$ as a $kG$-module.

\begin{proof} Recall that ${\mathrm{Res}}_N^{G}U=S_1^{a_1}\oplus \cdots \oplus S_r^{a_r}$, where $S_i^{a_i}$ is the unique largest $kN$-submodule that is isomorphic to a direct sum of copies of $S_i$. If $g\in G$, then $g(S_i^{a_i})$ is a direct sum of simple module $g(S_i)$, where $g(S_i)=S_j$. So $g(S_i^{a_i})\subseteq S_j^{a_j}$. It deduces that

$$U=gU=g(S_1^{a_1})\oplus \cdots \oplus g(S_r^{a_r}).$$

By counting dimension, one can check $g(S_i^{a_i})=S_j^{a_j}$. So $G$ permutes $S_1^{a_1},\cdots ,S_r^{a_r}$. Since ${\sum }_{g\in G}g(S_1^{a_1})$ is a $G$-invariant subspace of $U$, there is $U={\sum }_{g\in G}g(S_1^{a_1})$. So i) and ii) hold. Furthermore, iii) is trivial. \end{proof}

Corollary

For a normal subgroup $N⊲G$, suppose $\chi \in \mathrm{Irr}(G)$ and $\theta \in \mathrm{Irr}(H)$. If $⟨\theta ,{\mathrm{Res}}_N^G(\chi )⟩=e\ne 0$, then

• ${\mathrm{Res}}_N^G(\chi )={\sum }_{g\in [G/T]}e{\theta }^g$, where $T={\mathrm{Stab}}_G(\chi )$;
• let $\phi \in \mathrm{Irr}(T)$ with $⟨\phi ,{\mathrm{Ind}}_N^T(\theta )⟩\ne 0$, then ${\mathrm{Res}}_N^T(\phi )=e\theta$ and $\chi ={\mathrm{Ind}}_T^G(\phi )$.

Corollary

Let $k$ be a field with $\mathrm{char}k=p$, and let $G$ be a finite group. Then the common kernel of the action of $G$ on all the simple $kG$-modules is $O_p(G)$, where $O_p(G)$ is defined in ^slozsm. Thus, the simple $kG$-modules are precisely the simple $k[G/O_p(G)]$-module (via $G\to O_p(G)$).

Remark. The following proof is wrong. I can’t be bothered to write down the correct version at the moment, so this will have to do for now.

\begin{proof} Let $H$ be the common kernel of the action of $G$ on all simple module, that is,

$$H=\{g\in G:gv=v\text{ for any simple }kG\text{-module }S\text{ and any }v\in S\}.$$

If $S$ is a simple $kG$-module, then ${\mathrm{Res}}_{O_p(G)}^G(S)$ is semisimple $k{O}_p(G)$-module. Recall that for $\mathrm{char}k=p$ and a $p$-group $O_p(G)$, the only simple module is the trivial module. So $O_p(G)$ acts trivially on $S$ and so $O_p(G)⩽H$.

Now it suffices to show $H$ is a normal $p$-subgroup. Assume that there exists $h\in H$ such that $p\nmid o(h)$. Then by Maschke’s theorem, ${\mathrm{Res}}_{⟨h⟩}^G(kG)$ is a semisimple $k⟨h⟩$-module. Then ${\mathrm{Res}}_{⟨h⟩}^G(kG)$ is a direct sum of ${\mathrm{Res}}_{⟨h⟩}^G(S)$ where $S$ is a simple $kG$-module. Since $h\in H$, $h$ acts trivially on $kG$. However, it is impossible because $kG$ is faithful. Hence $H$ is a $p$-group. Since $H$ is the common kernel, it is a normal $p$-subgroup and so $H⩽O_p(G)$. \end{proof}

Theorem

Let $k$ be an algebraical closed field. If $G$ is a finite abelian group, then every simple $kG$-module has dimension $1$.

\begin{proof} See ^x79f5k. \end{proof}

Theorem

Let $k$ be a field with characteristic $p$, and let $G$ be a $p$-group. Then the regular module $kG$ is an indecomposable projective $kG$-module, that is, the projective cover of the trivial module. Every finitely generated projective $kG$-module is free. The only idempotent elements in $kG$ are $0$ and $1$.

\begin{proof} Recall that $kG={\oplus }_{\text{simple }S}{P}_S^{n_S}$, where $n_s={\dim }_{D}S$ and $D={\mathrm{End}}_{kG}(S)$. Since $kG$ is indecomposable, $kG=P_k^{n_k}$, where $k$ is the trivial $kG$-module. Note that ${\mathrm{End}}_{kG}(k)\subseteq {\mathrm{End}}_k(k)=k$, thus ${\mathrm{End}}_{kG}(k)=k$. Hence $n_k={\dim }_{k}k=1$, and then $kG$ is the projective cover of the trivial module.

For any finitely generated projective $kG$-module $P$, $P$ is a direct sum of indecomposable projective modules, and it deduces that $P\simeq (kG{)}^m$ and so $P$ is free.

For any idempotent $e\in kG$, there is $kG=kGe\oplus kG(1-e)$, where $kGe$ and $kG(1-e)$ are projective $kG$-module. As $kG$ is indecomposable, $e=0$ or $1$. Now we finish the proof. \end{proof}

The $p$-power radical modulo the commutator subspace

Definition

Let $A$ be a finite-dimensional $k$-algebra with $\mathrm{char}k=p$. Write $S=\mathrm{span}\{ab-ba:a,b\in A\}$ as a $k$-vector space and $T=\{r\in A:r^{p^n}\in S\text{ for some }n>0\}$.

Lemma

$T$ is a $k$-subspace of $A$ containing $S$.

\begin{proof} Firstly, note that $(a+b{)}^p\equiv a^p+b^p(\mathrm{mod}S)$ for any $a,b\in A$, because for any word $ababb\cdots a$, there are exactly $p$ elements that are equivalent in $A/S$. It deduces that $T$ is a $k$-space.

Next, if $a,b\in A$, then

$$(ab-ba{)}^p=(ab{)}^p-(ba{)}^p=a(b(ab{)}^{p-1})-(b(ab{)}^{p-1})a\in S$$

and so $S\subseteq T$.

Now we finish the proof. \end{proof}

Lemma

Let $A={\mathrm{M}}_n(k)$ and $\mathrm{char}k=p$. Then $S=T=\{M\in A:\mathrm{tr}(M)=0\}$.

\begin{proof} Define $S_0=\{M\in A:\mathrm{tr}(M)=0\}$. Note that $\mathrm{tr}(ab-ba)=0$ for any $a,b\in A$, so $S\subseteq S_0$.

On the other hand, the matrix $E_{ij}$ satisfies $E_{ij}=E_{ik}{E}_{kj}-E_{kj}{E}_{ik}$. So $E_{ij}\in S$ and so $S=S_0\subseteq T\subseteq A$. Since $S_0=\ker (M\mapsto \mathrm{tr}M)$, we have

$$\dim S_0+\dim (\mathrm{im}(\mathrm{tr}))=\dim A$$

and so $\dim S=\dim A-1$. As $T\ne A$, we have $S=T$. \end{proof}

Proposition

Let $A$ be a finite-dimensional $k$-algebra with $\mathrm{char}k=p$. Assume that $k$ is a splitting field for $A$, then the number of non-isomorphic simple $A$-modules equals $\dim A-\dim T$.

\begin{proof} Write $S(A)$, $T(A)$, $S(A/J(A))$, $T(A/J(A))$ for the construction $S,T$ applied to $A$ and $A/J(A)$. Since $J(A)$ is nilpotent, we have $J(A)\subseteq T(A)$. Also $(S(A)+J(A))/J(A)=S(A/J(A))$. (exercise)

We claim that $T(A)/J(A)=T(A/J(A))$. If $a\in A$ with $a^{p^n}\in S(A)$, then $(a+J(A){)}^{p^n}\in (S(A)+J(A))/J(A)=S(A/J(A))$. Hence $a+J(A)\in T(A/J(A))$. On the other hand, if $(a+J(A){)}^{p^n}\in S(A/J(A))$, then $a^{p^n}\in S(A)+J(A)\subseteq T(A)$. Now we proved the claim.

Recall that $A/J(A)$ is a direct sum of matrix algebras over $k$, and $S,T$ preserve direct sum. Remark that ${\mathrm{M}}_n(k)$ is simple, and by ^8o0awn, $\dim {\mathrm{M}}_n(k)-\dim T({\mathrm{M}}_n(k))=1$ is exactly the number of non-isomorphic simple ${\mathrm{M}}_n(k)$-modules. By ^f43ojz, the numbers of non-isomorphic simple $(A/J(A))$-modules equals

$$\dim (A/J(A))-\dim (T(A/J(A)))=\dim A-\dim T.$$

Now we finish the proof. \end{proof}

Lemma

Let $k$ be a field with $\mathrm{char}k=p$. Consider $A=kG$, then $S$ is the set of elements with the property that the sum of coefficient form each conjugacy class of $G$ is zero.

\begin{proof} Note that

$$S={\mathrm{span}}_k\{ab-ba:a,b\in kG\}={\mathrm{span}}_k\{gh-hg:g,h\in G\}$$

and $gh-hg=h(g^h)-hg$. \end{proof}

Definition

Let $p$ be a prime. An element of a finite group is called

• $p$-regular if it has order coprime to $p$;
• $p$-singular if it has order a power of $p$.

Brauer

Let $G$ be a finite group, and let $k$ be a splitting field for $G$ with $\mathrm{char}k=p$. The number of non-isomorphic of simple $kG$-modules equals the number of conjugacy classes of $p$-regular elements in $G$.

\begin{proof} By ^e69o14, we aim to show if $\{x_1,\cdots ,x_r\}$ is a set of representatives of conjugacy classes of $p$-regular elements of $G$, then $x_1+T,\cdots ,x_r+T$ is a $k$-basis of the $k$-space $kG/T$.

Let $x\in G$. One can write $x=st$, where $s$ is $p$-regular, $t$ is $p$-singular and $st=ts$. Then we have

$$(st-s{)}^{p^n}\equiv s^{p^n}{t}^{p^n}-s^{p^n}\equiv 0(\mathrm{mod}s)$$

for sufficiently large $n$. Hence $st-s\in T$ and so $s+T=st+T$. Thus $x_1+T,\cdots ,x_r+T$ spans $kG/T$.

Finally, show $x_1+T,\cdots ,x_r+T$ are linearly independent. Suppose ${\sum }_i{\lambda }_i{x}_i\in T$, then $({\sum }_i{\lambda }_i{x}_i{)}^{p^n}\in S$. Note that

$${\left(\sum _i{\lambda }_i{x}_i\right)}^{p^n}\equiv \sum _i{\lambda }_i^{p^n}{x}_i^{p^n}\equiv 0(\mathrm{mod}S),$$

so ${\sum }_i{\lambda }_i^{p^n}{x}_i^{p^n}\in S$. For sufficiently large $n$, $x_i^{p^n}=x_i$, so ${\sum }_i{\lambda }_i^{p^n}{x}_i\in S$. By ^wnm6ug, $x_1,\cdots ,x_n$ are linearly independent in $kG/S$. Then ${\lambda }_i^{p^n}=0={\lambda }_i$. Now we finish the proof. \end{proof}

$p$-modular system

Definition

Let $p$ be a prime. A $p$-modular system is a triple $(F,R,k)$ where $F$ is a field of character $0$ equipped with a discrete valuation, $R$ is the valuation ring of $F$ with maximal ideal $(\pi )$, and $k=R/(\pi )$ is the residue field of $R$ with $\mathrm{char}k=p$.

Given a finite group $G$, if both $F$ and $k$ are splitting for $G$, then we say that $(F,R,k)$ is a splitting $p$-modular system for $G$. This happens when $F$ contains a primitive $\exp (G)$th root of unity.

Lemma

Let $R$ be a DVR with maximal ideal $(\pi )$ and $k=R/(\pi )$. Let $G$ be a finite group.

• If $S$ is a simple $RG$-module, then $\pi S=0$.
• The simple $RG$-module are exactly the simple $kG$-modules made into $RG$-module via the surjection $RG\to kG$.
• For each $RG$-module $U$, $\pi U\subseteq \mathrm{Rad}U$. In particular, $\pi RG\subseteq J(RG)$.
• For each $RG$-module, we have $\mathrm{Rad}U/\pi U=\mathrm{Rad}(U/\pi U)$.

\begin{proof} i) Since $\pi S$ is a $RG$-submodule of $S$ and $S$ is simple, we know $\pi S=0$ or $S$. As $R$-modules, $\mathrm{Rad}(S)=J(R)S=(\pi )S\ne S$ by Nakayama lemma, so $\pi S=0$.

ii) Note that $kG=RG/(\pi G)$ and $\pi G$ annihilates simple $RG$-modules.

iii) If $V$ is a maximal submodule of $U$, then $U/V$ is simple. So $\pi (U/V)=0$ and $\pi U\subseteq V$. Hence $\pi U\subseteq \mathrm{Rad}U$.

iv) Recall that $\mathrm{Rad}U$ is the intersection of kernels of all morphisms from $U$ to simple modules. \end{proof}