$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Aut{Aut}$Let $f\in \Hom((\Bbb Z/p\Bbb Z)^2,\GL_{4}(\Bbb Z / p\Bbb Z))$ be an injective homomorphism. What is the number of isomorphism classes of semidirect products of the form $$(\Bbb Z/p\Bbb Z)^{4}\rtimes_{f} (\Bbb Z/p\Bbb Z)^2?$$ This allows us to compute the number of conjugacy classes of elementary abelian subgroups of order $p^2$ in $\GL_{4}(\Bbb Z / p\Bbb Z)=\Aut((\Bbb Z/p\Bbb Z)^{4})$ which is in fact not easy.
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4$\begingroup$ The title seems distinct from the question (even if there is a relation). If there is a reduction from the main question to the title question, please incorporate in the question. $\endgroup$– YCorCommented Nov 6, 2020 at 19:30
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3$\begingroup$ I am not sure that I believe that knowing just the number of isomorphism classes of semidirect products will enable you to compute the number of conjugacy classes of subgroups $C_p^2$ in ${\rm GL} (4,p)$. Not all of the homomorphisms $f$ are injective. Also, it is not clear (to me) that the isomorphism of two of the semidirect products would imply the conjugacy of the corresponding subgroups of ${\rm GL} (4,p)$. $\endgroup$– Derek HoltCommented Nov 6, 2020 at 19:50
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$\begingroup$ @Derek Holt, thank you for your comment. Of course f is injective. $\endgroup$– Nourddine SnanouCommented Nov 6, 2020 at 19:59
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1$\begingroup$ If it's any help, the answers (to both questions) when $p=2,3,5,7$ are $6,10,12,12$. That's as far as I can go with naive computation. $\endgroup$– Derek HoltCommented Nov 6, 2020 at 20:03
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2$\begingroup$ Note that for $p=2,3$ there are elements of order $p^2$ too. $\endgroup$– YCorCommented Nov 6, 2020 at 21:14
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