# Semi direct product to form a non abelian group of order $p^4$

I wish to ask following question. Kindly help me in following:

Let $p$ be an odd prime and $C_p$ denotes the cyclic group of order $p$. Let $G_1$ and $G_2$ be two groups both are isomorphic to $C_p \times C_p$. I wish to form a semi direct product (which is not a direct product) of $G_1$ and $G_2$ in which $G_1$ is normal so that I can produce a non abelian group of order $p^4$. To do this I need to define a non trivial homomorphism $\varphi:G_2 \rightarrow Aut(G_1)$. We know that $Aut(G_1) \cong GL(2,p)$ ($2 \times 2$ invertible matrices over the field $\mathbb{F}_p$). I know one non trivial homomorphism $\varphi: G_2 \rightarrow GL(2,p)$ which is defined as follows:

$(1,0) \mapsto \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$ and $(0,1) \mapsto \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$.

This will give me one non trivial semi direct product of $G_1$ and $G_2$.

I want to ask, Is there any other non trivial homomorphism $\varphi$ which gives me a non abelian group of order $p^4$ which is not isomorphic to what I have produced?

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The uniqueness of the nonabelian semidirect prodcut follows from the fact that a Sylow $p$-subgroup of ${\rm GL}(2,p)$ has order $p$. – Derek Holt Sep 1 '13 at 12:37

This is not an answer, but too long for a comment : There are two facts that help to tell if two semi-direct products are isomorphic : Let $H$ and $K$ be groups.
1. Let $\tau : K \to Aut(H)$ be a homomorphism, and $\sigma : K\to K$ an automorphism. Then $$H\times_{\tau} K \cong H\times_{\tau\circ\sigma} K$$ [Just take $f(h,k) = (h,\sigma(k))$ to be the isomorphism.]
2. Let $\tau_1$ and $\tau_2$ be monomorphisms from $K$ to $Aut(H)$ with $\tau_1(K) = \tau_2(K)$, then $$H\times_{\tau_1} K \cong H\times_{\tau_2} K$$ [Take $\sigma = \tau_2\circ\tau_1^{-1}$ in 1.]