# $G\rtimes_\phi H \cong G\rtimes_\psi H$ when certain automorphisms exist

I'm reviewing for exams and came across this problem from an older exam:

Let $G$ and $H$ be groups and let $\phi,\psi : H\to \operatorname{Aut}(G)$ be homomorphisms from $H$ into the group of automorphisms of $G$. If there exists $f\in \operatorname{Aut}(H)$ and $a\in\operatorname{Aut}(G)$ such that $\psi(f(h))=a\phi(h)a^{-1}$ as automorphisms of $G$ for all $h\in H$, then $G\rtimes_\phi H \cong G\rtimes_\psi H$.

I've proved this (unless I've made a mistake) by explicitly defining a map $F:G\rtimes_\phi H \to G\rtimes_\psi H$ by $(g,h)\mapsto(a(g),f(h))$ and checking that it is an isomorphism. However, choosing the map mostly involved guess-and-checking. That left me feeling like I somehow didn't do it "properly" and that I might have missed something I was supposed to notice or understand. I realize this is somewhat subjective, but I'm a bit suspicious, because the way I went about it makes the problem just seem like a strange, tedious way to test that someone knows the definition of a semidirect product.

Is there some other way to go about this problem (maybe using some clever argument about conjugation by $a$ inside the automorphism group?) that somehow illuminates or suggests something about the structure going on, or should it otherwise be obvious in some way what the isomorphism is? I guess my question is: is there something deeper going on here?

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I think that eventually you have to get down to defining the map explicitly and checking that it is an isomorphism.

But, conceptually: in general, an automorphism of $H$ will not induce an automorphism of $G\rtimes_{\phi} H$, because the action of $f(h)$ need not be related to the action of $h$ (of course, you get a different semidirect product, $G\rtimes_{\phi\circ f}H$, which may or may not be isomorphic). The automorphism $a$ provides the necessary "correction" by having $f(h)$ act on $a^{-1}(g)$ (instead of on $g$), so that the action of $f(h)$ on $a^{-1}(g)$ agrees with the original action of $h$ on $g$.

Intuitively: $H$ used to act on $G$ "in English", but after applying $f$ it is now trying to act "in Russian"; $a$ translates from Russian to English, so $(a^{-1}(g))^{f(h)}$ now makes sense, because $a^{-1}(g)$ is "in Russian", which $f(h)$ understands; then you translate the result back into English by applying $a$. It works exactly like a change-of-basis automorphism works for linear transformations. Interpreted that way, $(g,h)\mapsto (a(g),f(h))$ makes sense (that is, seems like the obvious thing to try), because we are simply performing the translations in both coordinates.

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Thanks. The conceptual comments and the analogy both help a lot. – matt May 9 '11 at 20:04

Both cases $G\rtimes_\phi H$ and $G\rtimes_\psi H$ are extensions of the group H by G. Extensions are classified by short exact sequences with in your case kernel G and image group H. The semi-direct products are exactly the so-called split extensions. To prove this the mentioned conjugation action appears naturally.

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Denoting $c_a$ for conjugation with $a$, the assumption means that the following diagram commutes (how can I draw nicer diagrams?): $$\begin{array}{rcl} H\quad & \stackrel{f}{\longrightarrow} &\quad H \\ {\scriptstyle\phi}\downarrow\ \ & &\ \downarrow{\scriptstyle \psi}\\ \text{Aut}(H) & \stackrel{c_a}{\longrightarrow} & \text{Aut}(H) \end{array}$$

where the horizontal arrows are group isomorphisms. As Arturo explained, this is similar to a 'change of basis' of the actions $\phi$ and $\psi$.

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There are better ways, but at least the horizontal arrows can be done with \stackrel. The rest, I sort of tried to align and size by hand. – Arturo Magidin May 9 '11 at 20:00