Why is $H \rtimes_{\varphi} K$ isomorphic to $H \rtimes_{\varphi \circ \phi} K$ where $\phi \in \text{Aut}(K)$?

Question

Why is $H \rtimes_{\varphi} K$ isomorphic to $H \rtimes_{\varphi \circ \phi} K$ where $\phi \in \text{Aut}(K)$?

I can see that $\varphi(K) = \varphi(\phi(K))$, but it is not clear to me how the semidirect products themselves are isomorphic. I tried constructing an isomorphism, but they were really messy and didn't seem to go anywhere.

Is there any easy way to see why this is true?

Answer 1

I'm going to use the usual notation, and write your groups as $H \rtimes_{\varphi} K$ (the normal subgroup doesn't get the extra bar on the product sign).

Let's write the general element of $H \rtimes_{\varphi} K$ as $kh$, where the product is given by $$ (k_1h_1)(k_2h_2) = (k_1k_2)(h_1^{\varphi(k_2)}h_2) $$

You can then define a map $f:H \rtimes_{\varphi} K\rightarrow H \rtimes_{\varphi \circ \phi} K$ $$ f(kh) = \phi^{-1}(k)h$$

This is clearly a bijection as a set map, so it just remains to show it's a homomorphism. We have \begin{align} f(k_1h_1k_2h_2) &= f(k_1k_2h_1^{\varphi(k_2)}h_2)\\ &= \phi^{-1}(k_1k_2)h_1^{\varphi(k_2)}h_2\\ &= \phi^{-1}(k_1)\phi^{-1}(k_2)h_1^{\varphi(k_2)}h_2 \end{align}

Meanwhile, in $H \rtimes_{\varphi \circ \phi} K$, \begin{align} f(k_1h_1)f(k_2h_2) &= \phi^{-1}(k)h_1\phi^{-1}(k)h_2\\ &= \phi^{-1}(k_1)\phi^{-1}(k_2)h_1^{\varphi \circ \phi\circ \phi^{-1}(k_2)}h_2\\ &= \phi^{-1}(k_1)\phi^{-1}(k_2)h_1^{\varphi(k_2)}h_2 \end{align}

which agrees with the above. So this is an isomorphism.