I have proved this proposition, which has been stated on other StackExchange pages.

Proposition: Let $N$ and $H$ be groups and let $\phi_1,\phi_2: H\to\newcommand{\Aut}{\operatorname{Aut}} \Aut(N)$ be group homomorphisms s.t. for some $g\in \Aut(N)$, $\phi_2(h)=g\phi_1(h)g^{-1}$ for all $h\in H$. Then $N \rtimes_{\phi_1} H \cong N\rtimes_{\phi_2} H$.

The map \begin{align*} f: N \rtimes_{\phi_1} H &\to N\rtimes_{\phi_2} H, (n,h)\mapsto (g(n),h). \end{align*} is an isomorphism.

Proof: This map is clearly well-defined and bijective since $g$ is a group automorphism. So it suffices to check the group homomorphism properties: $f((n,h)(n',h'))=f((n\phi_1(h)(n'),hh'))= (g(n\phi_1(h)(n')),hh')= (g(n) g\phi_1(h)(n'),hh')\\ =(g(n) g \phi_1(h)g^{-1}g(n'),hh')\\ =(g(n) \phi_2(h) g(n'),hh')\\ =(g(n),h)(g(n'),h')\\ = f(n,h)f(n',h').$ $ f((n,h)^{-1})=f((\phi_1(h^{-1})(n^{-1})),h^{-1})=(g\phi_1(h^{-1})(n^{-1}),h^{-1})\\ =((g\phi_1(h^{-1})g^{-1})(g(n^{-1})),h^{-1})\\ =(\phi_2(h^{-1})g(n^{-1}),h^{-1})\\ =(g(n),h)^{-1}\\ =f((n,h))^{-1}.$

Interestingly, when I do my homework, classifying groups of order 18,70,75, the converse is true.

For example, group of order 70.

"By Sylow Theorem, $n_5=n_7=1$. Then $N=N_5N_7 \cong Z/35Z$ is a normal subgroup since it is a product of two normal ones. Let $H$ be the $2-$Sylow subgroup, then $H=Z/2Z$. So we need to determine all the maps $\phi:Z/2Z \to \Aut(N) =\Aut(Z/5Z \times Z/7Z) \cong Z/4Z \times Z/6Z$. So we have the possibilities $\phi(1)=(0,0),(0,3),(2,0),(2,3)$. So there arev at most 4 groups of order $70$."

And according to the wiki, there are 4 groups of order 70, so the result follows.

My questions:

1) Is the converse true? Is there a counterexample? 2) Is my proof correct? THis is a ridiculously strong proposition, but I think my proof is not wrong.

  • 1
    $\begingroup$ Your condition isn't strong enough. You need for your two automorphisms to be conjugate in $\operatorname{Out}(G)$, so conjugate modulo the inner automorphisms, rather than equal modulo the inner automorphisms (which is the condition you have written). So you can cook up a counter-example to your condition by understanding my above sentence. $\endgroup$ – user1729 Nov 18 '19 at 16:40
  • 1
    $\begingroup$ (The condition I give above isn't equivalent either, at least not for infinite groups. Can't think of a counter-example at the moment though.) $\endgroup$ – user1729 Nov 18 '19 at 16:43
  • 1
    $\begingroup$ Weird stuff happens in infinite groups: math.stackexchange.com/questions/527800/… $\endgroup$ – AnalysisStudent0414 Nov 18 '19 at 16:43
  • $\begingroup$ let me write my proof for that, it is ridiculously strong, yes, But I do not see any mistakes $\endgroup$ – T C Nov 18 '19 at 16:43
  • $\begingroup$ @chi See my comment above. Your condition can be made much stronger. $\endgroup$ – user1729 Nov 18 '19 at 16:55

Consider the special case where $\phi_2$ is constant equal to the identity, so that one of the semi-direct products is actually a direct product. The converse of your proposition in that case would be :

If $N \rtimes_{\phi_1} H \cong N\times H$ then there is some $g\in \operatorname{Aut}(N)$ such that $\phi_2(h)=g\phi_1(h)g^{-1}$ for all $h\in H$, which in turn implies that $\phi_1$ is constant equal to the identity as well.

But it is well-known that non-trivial actions can lead to direct products.

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