Order of abelian subgroup of the automorphism group of an abelian group

Question

Suppose $A$ is a finite abelian group and $B\leq\operatorname{Aut}(A)$ is abelian. Is it possible for the order of $B$ to be strictly greater than the order of $A$? What if I additionally impose that the orders of $A,B$ are coprime?

As a special case, suppose $p$ is prime. Can $\operatorname{GL}_n(\mathbf F_p)$ have an abelian subgroup of order greater than $p^n$ (not divisible by $p$)? In other words, can an abelian group $B$ of order greater than $p^n$ have an $n$-dimensional faithful representation over a field with $p$ elements?

Edit: I just realised that the reasoning below is a bit silly, as $\lvert B\rvert<p^{n^2}$ is trivial, since $\lvert \operatorname{GL}_n(\mathbf F_p)\rvert<p^{n^2}$...

Below, I make some observations about this last case. Suppose $B$ is an abelian group admitting such a representation and $p$ does not divide the order of $B$.

Given such a representation, we can extend it to a representation over the algebraic closure of the $p$-element field and then decompose it into $n$ one-dimensional irreps. The image of a one-dimensional representation will be isomorphic to a finite subgroup of the multiplicative group of the field, and hence finite, generated by a root of unity.

This root of unity will be an eigenvalue of a matrix in the image of the original representation, and hence it is a root of a polynomial of degree $n$ with coefficients in $\mathbf F_p$. From this it follows that it is a primitive $d$-th root of unity for $d$ satisfying $p^m\equiv 1\pmod d$ for some $m\leq n$. This implies that $d<p^n$, which gives a weak bound of $\lvert B\rvert< p^{n^2}$, but I suspect it's not hard to get something much stronger.

Answer 1

As Nick Gill points out, one can certainly have $|B| > |A|$ if you do not assume coprimality. If you assume coprimality then yes $|B| < |A|$.

If $A_p$ is the Sylow $p$-subgroup of $A$ then $\DeclareMathOperator{Aut}{Aut}$$\Aut(A) \cong \prod_p \Aut(A_p)$ and $B$ is contained in the direct product of its projections to the factors $\Aut(A_p)$, so without loss of generality $A$ is a $p$-group and $B$ is a $p'$-subgroup of $\Aut(A)$.

The Frattini subgroup of $A$ is $pA$ and $A/pA \cong \mathbf{F}_p^n$ for some $n$ (we are writing $A$ additively). Therefore we have a natural homomorphism $\DeclareMathOperator{GL}{GL}$$\Aut(A) \to \GL_n(p)$. The kernel is a $p$-group! Indeed, if $\alpha$ is in the kernel then $\alpha = 1 + f$ for some homomorphism $f : A \to pA$. Suppose $p^k A = 0$, where $k \ge p$. Then, by the binomial theorem, $$\alpha^{p^k} = \sum_{i = 0}^{p^k} \binom{p^k}{i} f^i = 1,$$ since $f^i = 0$ for $i \ge k$ and $p^k$ divides $\binom{p^k}{i}$ for $0 < i < k$. Now since $B$ is a $p'$-group it follows that $B$ injects into $\GL_n(p)$, and its image there has order at most $p^n-1$, by the second comment of Nick Gill.

To explain this last part (in one way of a few), first suppose $V = \mathbf{F}_p^n$ is an irreducible $B$-module. Since the $1$-eigenspace of any element $b \in B$ is a $B$-submodule, this implies that $B$ acts freely on the nonzero elements of $V$, whence $|B| \le p^n-1$. In general, $V$ is completely reducible by Maschke's theorem, so we may write $V = V_1 \oplus \cdots \oplus V_k$ for some irreducible $B$-modules $V_i$. By the irreducible case, the image of $B$ in $\Aut(V_i)$ has order at most $|V_i|-1$, so the result follows.

Answer 2

Claim) Suppose $A$ is a finite non-trivial Abelian group. Let $B$ be an Abelian subgroup of ${\rm{Aut}}(A)$. Then, $|B|\leq |A|-1$ if $|A|$ and $|B|$ are coprime.

An elementary proof by induction that does not use representation theory

Alternative 1) There is a proper, non-trivial subgroup $M$ of $A$ preserved by every automorphism $f\in B$. Then any element $f$ of $B$ induces an automorphism of $M$, thus a homomorphism $B\rightarrow{\rm{Aut}}(M)$. The kernel is $\{f\in B\mid f\restriction_M={\rm{id}}\}$, hence an injection $B\big/\{f\in B\mid f\restriction_M={\rm{id}}\}\hookrightarrow {\rm{Aut}}(M)$. On the other hand, the homomorphism $B\rightarrow{\rm{Aut}}\left(\frac{A}{M}\right)$ (well defined since elements of $B$ preserve $M$) restricts to an injection on $\{f\in B\mid f\restriction_M={\rm{id}}\}$ due to $\gcd(|A|,|B|)=1$: Suppose $f\in B$ is an automorphism $A\rightarrow A$ which keeps every element of $M$ fixed, and induces identity on $\frac{A}{M}$. Pick generators $x_1+M,\dots,x_k+M$ of $\frac{A}{M}$. As the map $\frac{A}{M}\rightarrow \frac{A}{M}$ induced by $f$ is identity, for any $1\leq i\leq k$, there exists $m_i\in M$ with $f(x_i)=x_i+m_i$. But then, given $f\restriction_M={\rm{id}}$, one has $f^{\circ n}(x_i)=x_i+nm_i$ for every $n\in\Bbb{Z}$. We conclude that the order of $f$, a divisor of $|B|$, should be a multiple of $\min\{n\in\Bbb{Z}\mid n>0,nm_i=0\,\,\forall 1\leq i\leq k\}$. But this last number is a divisor of $|A|$ (recall that $m_1,\dots,m_k$ were elements of the additive Abelian group $A$). Hence the order of automorphism $f$ must be one. Consequently, aside from the injection $B\big/\{f\in B\mid f\restriction_M={\rm{id}}\}\hookrightarrow {\rm{Aut}}(M)$, we also have $\{f\in B\mid f\restriction_M={\rm{id}}\}\hookrightarrow{\rm{Aut}}\left(\frac{A}{M}\right)$. Applying the induction hypothesis twice to smaller groups $M$ and $\frac{A}{M}$: $$ |B|=\left|B\big/\{f\in B\mid f\restriction_M={\rm{id}}\}\right|\cdot\left|\{f\in B\mid f\restriction_M={\rm{id}}\}\right|\leq (|M|-1)\cdot\left(\left|\frac{A}{M}\right|-1\right)<|A|-1. $$

Alternative 2) No proper non-trivial subgroup of $A$ is preserved by $B$. In this case, one can use an argument similar to Sean Eberhard's: For each $f\in B\setminus\{{\rm{id}}\}$, the proper subgroup ${\rm{Fix}}(f)$ of $A$ is preserved by all elements of $B$ because $B$ is assumed to be Abelian. Therefore, ${\rm{Fix}}(f)$ must be trivial. That is, the action of $B$ on $A\setminus\{0\}$ is free. This implies $|B|\leq |A|-1$.

Note) The argument above shows that equality in $|B|\leq |A|-1$ is achieved iff $B\leq{\rm{Aut}}(A)$ acts freely and transitively on $A\setminus\{0\}$ (the inequality was strict in Alternative 1). In such a situation, all non-identity elements of $A$ must be of the same order. Thus $A$ must be an elementary Abelian group, that is of the form $\left(\frac{\Bbb{Z}}{p\Bbb{Z}}\right)^{\ell}$ for a prime $p$. One example would be to identify $\left(\frac{\Bbb{Z}}{p\Bbb{Z}}\right)^{\ell}$ with the additive group of the finite field $\Bbb{F}_{p^\ell}$, pick a generator of $\Bbb{F}_{p^\ell}^{\times}$ and consider its action on $A=\left(\Bbb{F}_{p^\ell},+\right)$ given by multiplication. This amounts to an Abelian subgroup of order $p^{\ell}-1$ of ${\rm{Aut}}(A)$. (I am not sure if these are the only examples.)