# Proving $H_1 \rtimes_{\theta_2}K \simeq H_1 \rtimes_{\theta_1}K$

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?

• 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 $p^3$. – Derek Holt Dec 22 '18 at 10:39
• 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 $K=C_p$, $H_1=C_p \times C_p$ ? – user401516 Dec 22 '18 at 10:41
• No, why should it mean that? – Derek Holt Dec 22 '18 at 10:54
• 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? – user401516 Dec 22 '18 at 11:01
• Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like. – Shaun Dec 22 '18 at 14:38