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Can someone help with the following problem?

Let $G$ be a non-abelian group of prime-power order $p^n$ and $E$ be an irreducible $G$-space over $\mathbb{C}$ giving a faithful representation of $G$.

(i) Show that $G$ has a normal abelian subgroup $H$ which properly contains $Z(G)$.

(ii) By complete reducibility, $E$ can be written as an $H$-direct sum $E=\bigoplus_{\psi}F_{\psi}$, where $F_{\psi}$ is the subspace of $E$ spanned by all 1-dimensional $H$-spaces with character $\psi$. Show that $E\neq F_{\psi}$ for any one $\psi$, using the facts that $Z(G)$ is a proper subgroup of $H$ and $E$ is faithful.

(iii) Note that $G$ permutes the subspaces $F_{\psi}$ transitively. Fix $\psi$ and let $F=F_{\psi}$. Let $H_1$ be the stabiliser of $F$ under the action of $G$. Clearly $H\subset H_1$. Show that $F$ is irreducible as an $H_1$-space ad that $E$ is isomorphic to the induced representation space $i_!F$.

(iv) What does (iii) imply about the (not necessarily faithful) irreducible representations of $G$?

For (i), I've proved that $Z(G)$ has order $p^k$ where $1\leq k\leq n-2$ and that any $H\leq G$ containing $Z(G)$ of order $p^{k+1}$ must be abelian; all that's left is to prove $H$ is normal in $G$. I can't quite get my head around (ii): what relevance does $Z(G)$ have to representations of $G$?

For (iii), since $H_1$ preserves $F$, $F$ is a representation of $H_1$, but why must it be irreducible? Proving it by inner products of characters doesn't look good; maybe we can prove that $H_1$ acts transitively on the 1-dimensional subspaces of $F$?

Many thanks for any help with this!

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2 Answers

up vote 2 down vote accepted

Here are few quick hints. To get a normal subgroup $H$ in (i), choose $H$ such that $H/Z(G)$ is a subgroup of $Z(G/Z(G))$.

In (ii), if $E = F_{\psi}$ for some $\psi$ then, since $F_{\psi}$ is the sum of equivalent 1-dimensional $H$-spaces, the action of $H$ on $F_{\psi}$ would be scalar which, since the representation is faithful, implies $H \le Z(G)$, contradiction.

For (iii), if $H_1$ is not irreducible, there is a nonzero proper $H'$-subspace $F' < F$, and then the $|G:H'|$ images of $F'$ under $G$ span a proper $G$-subspace of $E$, contradicting irreducibility.

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Thanks for this! I've now managed to solve the rest of the question too. But since your answer and Andreas Caranti's are so similar, I'm not sure which to set as my accepted answer! –  Harry Macpherson Apr 10 '13 at 12:31
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Some partial answers, possibly more later.

(i) is true in any non-abelian $p$-group. Take a normal subgroup $H/Z(G)$ of order $p$ of $G/Z(G)$, which exists as $Z(G/Z(G)) \ne 1$, then $H$ is normal in $G$ and abelian.

(ii) Consider the action of $H$ on a single $F_{\psi}$. This is given by $$ h v = \psi(h) v, $$ for all $v \in F_{\psi}$, that is, $H$ acts by scalar matrices on $F_{\psi}$. Now scalar matrices form the centre of $\operatorname{GL}(E)$, and $G$ is (isomorphic to) a subgroup of it, so if $E = F_{\psi}$, this means $H$ is central.

(iii) First, why does $G$ permutes the $F_{\psi}$. This is because if $v \in F_{\psi}$, and $g \in G$, $h \in H$ $$ h g v = g (g^{-1} h g)v = g \psi(g^{-1} h g) v = \psi(g^{-1} h g) g v, $$ as $H$ is normal in $G$. Thus $$ g F_{\psi} \subseteq F_{\psi'}, $$ where $\psi'(h) = \psi(g^{-1} h g)$.

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Thanks for this! I should have thought of the lattice isomorphism theorem for (i). I did know why $G$ permutes the $F_{\psi}$ though - that's why I put 'note that' rather than 'show that'! :-) –  Harry Macpherson Apr 10 '13 at 12:29
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