$N\rtimes_{\phi}H\cong N\rtimes_{\phi\circ\psi}H$ for $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, $\psi \in Aut(H)$

Question

Let $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, if $\psi \in Aut(H)$, prove that $$N\rtimes_{\phi}H\cong N\rtimes_{\phi\circ\psi}H.$$

This is mentioned in the original post. I am seeking proof of the statement.

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Edit: I have changed the order of $\phi$ and $\psi$ and the composition acts from right to left. Sorry about the vagueness.

Define $\varphi: N\rtimes_\phi H\to N\rtimes_{\phi\circ\psi}H$ by $$ (n,h)\mapsto (n,\psi^{-1}(h)) .$$ So it suffices to show that $\varphi$ is a group homomorphism.

\begin{align} \varphi((n_1,h_1)\cdot_\phi(n_2,h_2))&=\varphi((n_1\phi(h_1)(n_2),h_1h_2))\\ &=(n_1\phi(h_1)(n_2),\psi^{-1}(h_1)\psi^{-1}(h_2)) \end{align} On the other hand, \begin{align} \varphi((n_1,h_1))\cdot_{\phi\circ\psi}\varphi((n_2,h_2))&=(n_1,\psi^{-1}(h_1))\cdot_{\phi\circ\psi}(n_2,\psi^{-1}(h_2))\\ &=(n_1\phi\circ\psi\circ\psi^{-1}(h_1)(n_2),\psi^{-1}(h_1)\psi^{-1}(h_2)) \end{align}

Answer 1

$\DeclareMathOperator{\Aut}{Aut}$$\renewcommand{\phi}{\varphi}$I apologise for using a different notation (composition left-to-right and homo/auto-morphisms written as exponents), but if I do it differently, I might easily do it wrong. May try and translate it to different notation if deemed useful.

In $H \ltimes_{\phi} N$ the operation is given by $$ (h_{1}, n_{1}) \cdot (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{h_{2}^{\phi}} n_{2}). $$ (So in my notation $h_{2}^{\phi} \in \Aut(N)$ is the image of $h_{2}$ under $\phi$, and $n_{1}^{h_{2}^{\phi}}$ denotes its operation on $n_{1}$.)

In $H \ltimes_{\psi \phi} N$ the operation is given by $$ (h_{1}, n_{1}) \circ (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{h_{2}^{\psi \phi}} n_{2}). $$ (Here $\psi \phi$ is the composition of $\psi$ first and $\phi$ second.)

Now an isomorphism between the two groups is given by \begin{align} G :\ &H \ltimes_{\phi} N \to H \ltimes_{\psi \phi} N\\ &(h, n) \mapsto (h^{\psi^{-1}}, n). \end{align} (Note the critical inverse, see the calculation below.)

In fact

$$ G((h_{1}, n_{1})) \circ G((h_{2}, n_{2})) = (h_{1}^{\psi^{-1}}, n_{1}) \circ (h_{2}^{\psi^{-1}}, n_{2}) = (h_{1}^{\psi^{-1}} h_{2}^{\psi^{-1}}, n_{1}^{h_{2}^{\psi^{-1} \psi \phi}} n_{2}) = ((h_{1} h_{2})^{\psi^{-1}}, n_{1}^{h_{2}^{\phi}} n_{2}), $$ which equals $$ G((h_{1}, n_{1}) \cdot (h_{2}, n_{2})) = G((h_{1} h_{2}, n_{1}^{h_{2}^{\phi}} n_{2})) = ((h_{1} h_{2})^{\psi^{-1}}, n_{1}^{h_{2}^{\phi}} n_{2}). $$

Answer 2

If you have two group structures on the same group, saying the identity is an isomorphism is saying the two group structures are actually the same. Here this couldn't possible be the case unless your automorphism is the identity.

Instead your map should be $(n,h) \rightarrow (n,\psi(h))$.