irreducible representation of non-abelian p-group

Question

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!

Answer 1

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.

Answer 2

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 $\psi'(h) = \psi(g^{-1} h g)$.