The rank of indecomposable finite abelian 2-group

Question

$\DeclareMathOperator\rank{rank}$Let $P$ be a finite $p$-group. The rank of $P$ is $\log_{p}|P/\Phi(P)|$ where $\Phi(P)$ is the Frattini subgroup of $P$, we write $\rank(P)=\log_{p}|P/\Phi(P)|$.

Let a finite group $A$ act faithfully on a finite $2$-group $G$.

An $A$-invariant subgroup $H$ is called a $A$-direct factor if there exists an $A$-invariant subgroup $K$ such that $G=H\times K$. If there are no proper nontrivial $A$-direct factors of $G$, then $G$ is said to be $A$-indecomposable (or indecomposable if $A=1$). The next $2$ results on the rank of $A$-indecomposable $2$-group are well-known.

• Assume $G$ is $A$-indecomposable abelian $2$-group. If $A=1$, then $\rank(G)=1$.

• Assume $G$ is $A$-indecomposable abelian $2$-group. If $\gcd(|A|,|G|)=1$, then $G$ is homocyclic (direct product of cyclic groups of the same order), also $\rank(G)$ is equal to the dimension of the irreducible $GF(2)[A]$-module $G/\Phi(G)$ ($GF(2)$ is the finite field of $2$ elements).

Question Assume $G$ is $A$-indecomposable abelian $2$-group. If $A$ is a cyclic finite $2$-group, then could we say something about $\rank(G)$?

Easy version Assume $G$ is $A$-indecomposable abelian $2$-group. If $A$ is a cyclic group of order $2$, then could we say something about $\rank(G)$?

Using linear algebra, one could determine $\rank(G)$ in Easy version when $G$ is elementary abelian, that is $\rank(G)=1$ or $2$.

Answer 1

No, the rank of $G$ could be arbitrary large. Here is a counter-example that works for all primes $p$. Fix a positive integer $n$, and let $G$ be the direct sum of $n$ copies of $\mathbb{Z}/p^2\mathbb{Z}$ and $x_1,\ldots,x_n$ be the canonical basis of $G$. Define the automorphism $\sigma$ of $G$ by $$\sigma x_i=x_i+px_{i+1} \quad \mbox{ for } 1\leq i\leq n-1,\quad \mbox{and}\quad \sigma x_n=x_n.$$ It follows immediately that $p(\sigma-1)=0$ and $(\sigma-1)^2=0$; hence $A=\langle \sigma \rangle$ has order $p$. It remains to see that $G$ is an indecomospsoble $A$-module. Suppose $H$ is a non-trivial direct factor of $G$, and let $H_i=H\cap \langle x_i,\ldots,x_n\rangle$. We claim that $H_i\neq 0$ for all indices $i$. Indeed, we have $H_1=H\neq 0$, and by induction if $H_i\neq 0$ for some $i<n$, then $H_{i+1}\neq 0$, as otherwise, by picking a non-zero element $x=\sum_{j\geq i} \alpha_jx_j$ in $H_i$, we find $$ (\sigma-1)x=p\alpha_j x_{j+1}+\cdots+p\alpha_{n-1}x_{n}=0.$$ Hence $p$ divides $\alpha_j$ for $j=1,\ldots,n-1$. It follows that $px=p\alpha_nx_n$ lies in $H_n=0$ (since $H_n\leq H_{i+1}=0$) and so $p$ divides $\alpha_n$. Now, by setting $$\alpha_j=p\beta_j \quad \mbox{for } j\geq i,\quad \mbox{and} \quad y=\sum_{j\geq i} \beta_jx_j,$$ we find $py\in H_i$. Writing $y=h+k$ with $h\in H$ and $k$ lies in some $A$-complement of $H$, we deduce that $pk=0$ and so $(\sigma-1)k=0$. It follows that $$(\sigma-1)y=(\sigma-1)h \in H.$$ As $(\sigma-1)y=\sum_{j= i}^{n-1}p\beta_jx_{j+1}$, the above implies $(\sigma-1)y\in H_{i+1}=0$. This shows that $p\beta_j=\alpha_j=0$ for $j=i,\ldots,n-1$; thus $x=\alpha_nx_n$ is a non-zero element in $H_n=0$, a contradiction; this proves the claim.

It follows that every $A$-invariant direct factor of $G$ meets $\langle x_n\rangle$ non-trivially; in particular, $G$ is $A$-indecomposable.

It wouldn't be difficult to modify the above example in order to obtain reacher families of indecomposable $A$-modules. Note also, if you are interested, the rank of $G$ is bounded in terms of the rank of the submodule of $A$-fixed points.