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$\DeclareMathOperator{\GL}{\operatorname{GL}}$Let $G$ be an elementary abelian $p$-group of rank $2$. Let $\alpha, \beta :G\rightarrow \GL_{n}(\Bbb Z / p\Bbb Z)$ be two injective homomorphisms. The homomorphisms $\alpha$ and $\beta$ are called similar if there exist $\sigma \in {\rm Aut}(A)$, $\rho \in{\rm Aut}(G)$ such that $(\beta \circ \rho )(g)=\sigma \circ \alpha (g)\circ \sigma^{-1}$ for all $g\in G$. Since $\alpha$ and $\beta$ are injective then $\alpha(G)$ and $\beta(G)$ are elementary abelian p-subgroups of rank 2 in $GL_{n}(\Bbb Z / p\Bbb Z)$.Thus, set $\alpha(G)=\langle A,B\rangle$ and $\beta(G)=\langle A',B'\rangle$ for some $A$ ,$B$ ,$A'$ and $B'\in \GL_{n}(\Bbb Z / p\Bbb Z)$.

The pairs $(A,B)$ and $(A',B')$ are called simultaneously similar if there is $U\in \operatorname{GL}_n(\mathbb F_p)$ such that $A'=UAU^{-1}$, $B'=UBU^{-1}$.

Are $\alpha$ and $\beta$ similar homomorphisms if and only if the pairs $(A,B)$ and $(A',B')$ are simultaneously similar?.

What is the number of simultaneous similarity classes of the pair $(A,B)$ in $GL_{n}(\Bbb Z / p\Bbb Z)$?.

Thank you in advance.

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    $\begingroup$ It seems implicit that $A,B$ are the images of two given generators of $G$ (rather than an arbitrary generating pair of $\alpha(G)$, as the current notation seems to indicate)? Also $A$ is used to denote a group, and then to denote a matrix. $\endgroup$
    – YCor
    Commented Nov 9, 2020 at 4:12
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    $\begingroup$ But what is $A$ in $\mathrm{Aut}(A)$. Probably not the image of the first generator, also denoted $A$. $\endgroup$
    – YCor
    Commented Nov 9, 2020 at 12:08
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    $\begingroup$ I am not sure whether I am understanding the definitions properly, and you need to sort out your notation, but it looks as thought the pairs $(A,B)$ and $(B,A)$ could fail to be simultaneously similar, but the corresponding homomorphisms $\alpha, \beta$ would be similar, because there is an automorphism of $\langle A,B \rangle$ that swaps $A$ and $B$. $\endgroup$
    – Derek Holt
    Commented Nov 9, 2020 at 13:36
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    $\begingroup$ I added some clarifications to the question. @YCor of course I m not loking to A as image of the first generator of G. In fact, if $\alpha$ and $\beta$ are similar it follows that $\alpha(G)$ and $\beta(G)$ are conjugate subgroup which led to the simultaneous similarity of their generators. $\endgroup$
    – Nourr Mga
    Commented Nov 9, 2020 at 21:58
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    $\begingroup$ So you're using $A$ for two different things; it seems that $A$ denotes $\mathrm{GL}_2(\mathbf{Z}/p\mathbf{Z})$ at the beginning, and the image of the first generator (hence a given element of $\mathrm{GL}_2(\mathbf{Z}/p\mathbf{Z})$) later. Why don't you fix it? $\endgroup$
    – YCor
    Commented Nov 9, 2020 at 21:59

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