Action of $\operatorname{Aut}(A) \times \operatorname{Aut}(G)$ on $\operatorname{Hom}(G,\operatorname{Aut}(A))$

Question

Let $G$ and $A$ be two finite abelian groups. Let $T:= \operatorname{Hom}(G,\operatorname{Aut}(A))$ be the set of all injective group homomorphisms from $G$ to $\operatorname{Aut}(A)$.

Take $g := (\sigma,\rho)\in \operatorname{Aut}(A) \times \operatorname{Aut}(G)$ and let $$T^g := \{\alpha \in T :\alpha \circ \rho = \gamma_{\sigma} \circ \alpha \}.$$ where $\gamma_{\sigma}$ denotes the conjugation by $\sigma$ in $Aut(A)$. Are there some conditions on $G$ and $A$, or examples, under which we can compute easily the cardinality of $T^g$?

Thank you very much.

Answer 1

Just some very partial work, that I couldn't accomodate into a comment. So:

\begin{alignat}{1} T^{(\sigma,\rho)} &:=\{\alpha\in T\mid \alpha(\rho(g))=\sigma^{-1}\alpha(g)\sigma, \space\forall g\in G\} \\ &=\{\alpha\in T\mid \sigma\alpha(\rho(g))=\alpha(g)\sigma, \space\forall g\in G\} \\ \tag 1 \end{alignat}

As special cases:

• $(\sigma,\rho)=(\iota_A,\iota_G)$

\begin{alignat}{1} T^{(\iota_A,\iota_G)} &=\{\alpha\in T\mid \alpha(g)=\alpha(g), \space\forall g\in G\} \\ &= T \tag 2 \end{alignat}

• $(\sigma,\rho)=(\iota_A,\rho\ne\iota_G)$

\begin{alignat}{1} T^{(\iota_A,\rho)} &=\{\alpha\in T\mid \alpha(\rho(g))=\alpha(g), \space\forall g\in G\} \\ \tag 3 \end{alignat}

• $(\sigma,\rho)=(\sigma\ne\iota_A,\iota_G)$

\begin{alignat}{1} T^{(\sigma,\iota_G)} &=\{\alpha\in T\mid \sigma\alpha(g)=\alpha(g)\sigma, \space\forall g\in G\} \\ &=\{\alpha\in T\mid \alpha(g)\in C_{\operatorname{Aut}(A)}(\sigma), \space\forall g\in G\} \\ &=\{\alpha\in T\mid \alpha(G)\le C_{\operatorname{Aut}(A)}(\sigma)\} \\ \tag 4 \end{alignat}

Answer 2

I might have misunderstood anything, but if by "action" in the title you mean the following:

\begin{alignat*}{2} \mathcal{A}:(\operatorname{Aut}(A)\times\operatorname{Aut}(G))\times T & \longrightarrow & T \\ ((\sigma,\rho),\alpha) & \longmapsto & (\sigma,\rho)\cdot\alpha: G & \longrightarrow \operatorname{Aut}(A) \\ && g &\longmapsto \sigma^{-1}\alpha(\rho(g))\sigma \\ \tag 0 \end{alignat*}

then seemingly it fails to fulfil the "Action#2" property for actions (see $(1)$ and $(2)$ hereunder). In that case, group actions wouldn't assist the matter.

---

• Good definition: for every $\sigma\in \operatorname{Aut}(A)$, $\rho \in\operatorname{Aut}(G)$ and $g,h\in G$, indeed $\sigma^{-1}\alpha(\rho(g))\sigma\in \operatorname{Aut}(A)$ and: \begin{alignat}{1} ((\sigma,\rho)\cdot\alpha)(gh) &= \sigma^{-1}\alpha(\rho(gh))\sigma \\ &= \sigma^{-1}\alpha(\rho(g)\rho(h))\sigma \\ &= \sigma^{-1}\alpha(\rho(g))\alpha(\rho(h))\sigma \\ &= \sigma^{-1}\alpha(\rho(g))\sigma\sigma^{-1}\alpha(\rho(h))\sigma \\ &= (\sigma^{-1}\alpha(\rho(g))\sigma)(\sigma^{-1}\alpha(\rho(h))\sigma) \\ &= ((\sigma,\rho)\cdot\alpha)(g)((\sigma,\rho)\cdot\alpha)(h) \\ \end{alignat} whence indeed $(\sigma,\rho)\cdot\alpha\in T$

• Action#1: $\forall \alpha\in T,\forall g\in G$, \begin{alignat}{1} ((\iota_A,\iota_G)\cdot\alpha)(g) &= \iota_A^{-1}\alpha(\iota_G(g))\iota_A \\ &= \alpha(g) \\ \end{alignat} whence $\forall\alpha\in T$: $$(\iota_A,\iota_G)\cdot\alpha=\alpha$$

• Action#2: $\forall \sigma,\sigma'\in\operatorname{Aut}(A),\forall \rho,\rho'\in\operatorname{Aut}(G), \forall \alpha\in T,\forall g\in G$, \begin{alignat}{1} (((\sigma,\rho)(\sigma',\rho'))\cdot\alpha)(g) &= ((\sigma\sigma',\rho\rho')\cdot\alpha)(g) \\ &= (\sigma\sigma')^{-1}\alpha((\rho\rho')(g))\sigma\sigma' \\ &= \sigma'^{-1}\sigma^{-1}\alpha(\rho(\rho'(g)))\sigma\sigma' \\ \tag 1 \end{alignat} while \begin{alignat}{1} ((\sigma,\rho)\cdot((\sigma',\rho')\cdot\alpha))(g) &= \sigma^{-1}(((\sigma',\rho')\cdot\alpha)(\rho(g))\sigma \\ &= \sigma^{-1}\sigma'^{-1}\alpha(\rho'(\rho(g))\sigma'\sigma \\ \tag 2 \end{alignat} Therefore, $\mathcal{A}$ defined in $(0)$ is not an action.

However, the map $\mathcal{A}$ becomes indeed an action if you narrow the group which acts on $T$ to $Z(\operatorname{Aut}(A))\times Z(\operatorname{Aut}(G))$, as in this case $(1)$ and $(2)$ match. In that case:

\begin{alignat}{1} T^{(\sigma,\rho)} &:=\{\alpha\in T\mid \alpha(\rho(g))=\sigma\alpha(g)\sigma^{-1}, \space\forall g\in G\} \\ &=\{\alpha\in T\mid \sigma^{-1}\alpha(\rho(g))\sigma=\alpha(g), \space\forall g\in G\} \\ &=\{\alpha\in T\mid ((\sigma,\rho)\cdot \alpha)(g)=\alpha(g), \space\forall g\in G\} \\ &=\{\alpha\in T\mid (\sigma,\rho)\cdot \alpha=\alpha\} \\ &= \operatorname{Fix}(\sigma,\rho) \\ \end{alignat}

So, for instance, if you'd know that the action is transitive (so it had one orbit, only), then you could conclude that (Burnside's Lemma):

$$\sum_{(\sigma,\rho)\in Z(\operatorname{Aut}(A))\times Z(\operatorname{Aut}(G))}|T^{(\sigma,\rho)}|=|Z(\operatorname{Aut}(A))\times Z(\operatorname{Aut}(G))|$$

(I realized just now that I didn't take into accout the fact that $A$ and $G$ are finite abelian groups, by assumption, which should simplify this all...)