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Let $K=C_p$ be a cyclic group of order $p$ (prime). Let $H_1 = C_p \times C_p$, and $\theta_1,\theta_2 : K \to Aut(H_1)$ two homomorphisms. Denote $G_1 = H_1 \rtimes_{\theta_1}K$ and $G_2 = H_1 \rtimes_{\theta_2}K$.

Assume $G_1, G_2$ are nonabelian. Prove: $G_1 \simeq G_2$


I tried to check what are the (non trivial) homomorphisms $\theta :C_p \to Aut(C_p \times C_p)$, but I can't find any useful property. Any idea how can I characterize those homomorphisms?

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    $\begingroup$ As stated this is not true, because $\theta_1$ could be trivial and $\theta_2$ nontrivial. It is true if you assume that $\theta_1$ and $\theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$. $\endgroup$ – Derek Holt Dec 22 '18 at 10:39
  • $\begingroup$ Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p \times C_p$ ? $\endgroup$ – user401516 Dec 22 '18 at 10:41
  • $\begingroup$ No, why should it mean that? $\endgroup$ – Derek Holt Dec 22 '18 at 10:54
  • $\begingroup$ First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' \rtimes K' = G_2$. What am I missing? $\endgroup$ – user401516 Dec 22 '18 at 11:01
  • $\begingroup$ Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like. $\endgroup$ – Shaun Dec 22 '18 at 14:38

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