This is a followup to this previous question: Do homomorphisms $H \to \operatorname{Aut}(K)$ that coincide at the level of $\operatorname{Out}(K)$ induce isomorphic semidirect products?

I am trying to understand why $S_n$ doesn't seem to appear as a normal subgroup of a nontrivial semidirect product except when $n=6$. Two relevant facts: $S_n$ is centerless when $n \geq 3$, and its outer automorphism group is trivial except when $n = 6$. This raises the more general question:

If $\phi,\psi:H \to \operatorname{Aut}(K)$ induce the same map $H \to \operatorname{Out}(K)$, and if $K$ is centerless, then is it true that $K \rtimes_\phi H \cong K \rtimes_\psi H$?

  • $\begingroup$ I don't know if this exactly answers your question, but I think what you can say is that $K \rtimes_\phi H \cong K \rtimes_\psi H$ if and only if the images of $H$ under $\phi$ and $\psi$ are conjugate in $\mathrm{Aut}(K)$. That is, when there is some $\alpha \in \mathrm{Aut}(K)$ so that $\psi(h) = \alpha^{-1} \circ \phi(h) \circ \alpha$ for every $h \in H$. $\endgroup$ – Nick Mar 26 '16 at 2:39
  • $\begingroup$ @Nick That is not a necessary consition for the semidirect prodcuts to be isomorphic. The only semidirect product i which $K=S_5$ and $H=C_2$ is $S_5 \times C_2$, but there are two classes of elements of order $2$ in $K$. $\endgroup$ – Derek Holt Mar 26 '16 at 9:15

The answer is still no.

As an example, let $K=A_5$ and $H = \langle a,b\mid a^4=b^2=1,ab=ba \rangle = C_4 \times C_2$.

Choose $\phi$ such that $\phi(a)$ is conjugation by $(1,2,3,4)$ and $\phi(b)=1$.

Choose $\psi$ such that $\psi(a)$ is conjugation by $(1,2)$ and $\psi(b)$ is conjugation by $(1,2)(3,4)$.

Then $\phi$ and $\psi$ induce the same map to ${\rm Out}(K)$, but the associated semidirect products are not isomorphic, essentially because $\phi$ and $\psi$ have different kernels. In fact the one with $\phi$ has centre a Klein $4$-group and the one with $\psi$ has cyclic centre, which is similar to the example I gave for the previous question.

  • $\begingroup$ Thanks again, Derek. Do you have any idea what would make $S_n$ special in the way I described above? It seems quite peculiar to me. $\endgroup$ – Alex Provost Mar 26 '16 at 15:39

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