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?

  • 1
    $\begingroup$ I still don't know much about semidirect products, but found Jyrki's answer to this question helpful in general. Maybe you're already aware of these things about homomorphisms out of semidirect products, though. $\endgroup$ – pjs36 Nov 12 '17 at 5:49

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.

  • $\begingroup$ That’s perfect, thanks! Also, thanks for pointing out the notation typo, fixed it. $\endgroup$ – jackson5 Nov 12 '17 at 8:02

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