# Semidirect Products of the form $(\mathbb{Z}_p \times \mathbb{Z}_p)\rtimes_{\phi}\operatorname{GL}(2,p)$

What are the different (non-isomorphic) semidirect products $(\mathbb{Z}_p \times \mathbb{Z}_p)\rtimes_{\phi}\operatorname{GL}(2,p)$, when $\phi \colon \operatorname{GL}(2,p)\rightarrow\operatorname{Aut}(\mathbb{Z}_p \times \mathbb{Z}_p)$ is injective homomorphism? ($p$ is any prime).

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Ehr... $GL(2,p)$ is the automorphism group of $\mathbb{Z}_p\times\mathbb{Z}_p$, so the only $\phi$s are the automorphisms of $GL(2,p)$ itself. All those are isomorphic, you are just looking at the holomorph of $\mathbb{Z}_p\times\mathbb{Z}_p$. – Arturo Magidin Apr 9 '11 at 4:30
@Arturo: But I could not convince that two different injective homomorphisms from $GL(2,p)$ to $Aut(Z_p\times Z_p)$ will give isomorphic semidirect products; the well known result is that if $H$ is cyclic, then any two monomorphisms $\psi_1, \psi_2\colon H\rightarrow Aut(N)$, s.t. $\psi_1(H)$ and $\psi_1(H)$ are conjugate, will give isomorphic semidirect products of $N$ by $H$; but in this example $H=GL(2,p)$ is not cyclic, I could not apply the criteria stated. – user8186 Apr 9 '11 at 4:35
@user9332: You don't need that. You are trying to compare $K\rtimes_{\phi}H$ and $K\rtimes_{\phi\circ\psi}H$ where $\psi\in\mathrm{Aut}(H)$. These two semidirect products are always isomorphic, no matter what $\phi$ is. – Arturo Magidin Apr 9 '11 at 4:44
@user9332: Note: an "injective homomorphism from $GL(2,p)$ to $\mathrm{Aut}(\mathbb{Z}_p\times\mathbb{Z}_p)$ Is just an automorphism of $\mathrm{Aut}(\mathbb{Z}_p\times\mathbb{Z}_p)$. The two groups are isomorphic and finite, so an injective homomorphism is just an automorphism. Perhaps this is what is tying you up in knots? – Arturo Magidin Apr 9 '11 at 4:48

The automorphism group of $\mathbb{Z}_p\times\mathbb{Z}_p$ is isomorphic to $\mathrm{GL}(2,p)$ (view $\mathbb{Z}_p\times\mathbb{Z}_p$ as a two-dimensional vector space over $\mathbb{F}_p$, the field of $p$ elements). So your $\phi$ is just an automorphism of $\mathrm{Aut}(\mathbb{Z}_p\times\mathbb{Z}_p)$.

If $\phi$ is the identity, then you have the holomorph of $\mathbb{Z}_p\times\mathbb{Z}_p$, where the holomorph of $G$ is $\mathrm{Hol}(G) = G\rtimes \mathrm{Aut}(G)$, with the obvious action.

And so the question comes down to the following: if $K\rtimes_{\phi} H$ is a semidirect product, and $\psi\colon H\to H$ is an automorphism, is $K\rtimes_{\phi\circ\psi}H$ isomorphic to $K\rtimes_\phi H$?

The answer is yes. Define $F\colon K\rtimes_{\phi}H\to K\rtimes_{\phi\circ \psi}H$ by $F(h,k) = (\psi^{-1}(h),k)$. This is trivially a bijection, so we just need to show that it is a group homomorphism.

If $(h,k),(y,x)\in K\rtimes_{\phi}H$, we have: \begin{align*} (h,k)(y,x) &= (hy, k^{\phi(y)}x).\\ F\bigl( (h,k)(y,x)\bigr) &= (\psi^{-1}(hy),k^{\phi(y)}x).\\ F(h,k)F(y,x) &= (\psi^{-1}(h),k)(\psi^{-1}(y),x)\\ &= \bigl( \psi^{-1}(h)\psi^{-1}(y), k^{\phi\circ\psi(\psi^{-1}(y))}x\bigr)\\ &= \bigl(\psi^{-1}(hy), k^{\phi(y)}x\bigr)\\ &= F\bigl( (h,k)(y,x)\bigr). \end{align*} So $F$ is a group homomorphism, hence an isomorphism.

Applying this to the particular case you have, there is exactly one such semidirect product, isomorphic to the holomorph of $\mathbb{Z}_p\times\mathbb{Z}_p$.

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