I recently realized that I don't know of any group that is a nontrivial semidirect product of some symmetric group $S_n$ and another group ($S_n$ being the normal subgroup), except when $n=6$. (For instance, $\operatorname{Aut}(S_6) \cong S_6 \rtimes C_2$.) I believe that this is because all outer automorphism groups of $S_n$ are trivial except when $n = 6$.

In fact I believe this stronger statement is true:

If $\phi,\psi:H \to \operatorname{Aut}(K)$ induce the same map $H \to \operatorname{Out}(K)$, then $K \rtimes_\phi H \cong K \rtimes_\psi H$.

Writing things out, the hypothesis means that for all $h \in H$, there exists some $k_0 \in K$ (possibly dependent on $h$) such that $\phi_h(k) = k_0\psi_h(k)k_0^{-1}$ for all $k \in K$. (So this essentially reduces to this unanswered question modulo the fact that $k_0$ varies with $h$.)

If our isomorphism $\theta:K \rtimes_\phi H \to K \rtimes_\psi H$ is to be of the form $\theta(k,h) = (\alpha(k),h)$ for some bijection $\alpha$, then for $\theta$ to be a homomorphism we must have equality between the expressions $$\theta((k,h)(k',h')) = \theta(k\phi_h(k'),hh') = (\alpha(k\phi_h(k')),hh')$$ and $$\theta(k,h)\theta(k',h') = (\alpha(k),h)(\alpha(k'),h') = (\alpha(k)\psi_h(\alpha(k')),hh'),$$

i.e., we must have $$\alpha(k\phi_h(k')) = \alpha(k)\psi_h(\alpha(k')) = \alpha(k)k_0^{-1}\phi_h(\alpha(k'))k_0.$$

I really don't see how we can choose such an $\alpha$. But maybe $\theta$ has a more complicated form. (Or maybe the statement is false!)

Update: As Derek Holt has shown below, the statement is false in this generality. What can we say if $Z(K) = 1$, i.e., $K \cong \operatorname{Inn}(K)$? This is the case for $S_n$ ($n \geq 3$), for instance.

  • 1
    $\begingroup$ The statement is probably false. Take $H=K$ nonabelian and consider the trivial homomorphism $\phi:K\to\mathrm{Aut}(K)$, $\phi(k)=1_K$ and the homomorphism $\psi:K\to\mathrm{Inn}(K)\hookrightarrow\mathrm{Aut}(K)$. I suspect is is not too hard to prove $$K\rtimes_\phi K=K\times K\ncong K\rtimes_\psi K$$ $\endgroup$ – David Hill Mar 23 '16 at 19:29
  • 1
    $\begingroup$ @David: in fact the second semidirect product is also isomorphic to the direct product. This often surprises people at first, but is well known and not hard to check. (In the direct product, take one of the copies as your normal subgroup, and the diagonal copy as the complement. The diagonal acts by conjugation on the first copy.) $\endgroup$ – verret Mar 23 '16 at 23:13
  • $\begingroup$ I think answer to this question may help. math.stackexchange.com/questions/3106299/… $\endgroup$ – Andrews Feb 16 at 9:13

It's not true in general. Let $K = \langle x,y \rangle$ be dihedral of order $8$ and $|H|=2$ with $H=\langle z \rangle$.

Let $\phi$ be the trivial homomorphism, so $K \rtimes_\phi H = K \times H$.

Let $\psi$ map $z$ to the inner automorphism of $K$ defined by conjugation by $x$.

Then $\phi$ and $\psi$ both induce the trivial map to ${\rm Out}(K)$. But $K \rtimes_\psi H$ is not isomorphic to $K \rtimes_\phi H$.

To see that, note the centre of the group using $\phi$ is a Klein $4$-group, whereas the centre of the one using $\psi$ is cyclic of order $4$ generated by $zx^{-1}$.

I think it possible that your conjecture holds when $Z(K)=1$, but would need to think more about it.

  • $\begingroup$ If you take $K=S_3$ you can do the same as above, so $Z(K)=1$ does not seem to help. $\endgroup$ – David Hill Mar 23 '16 at 20:07
  • 2
    $\begingroup$ I don't think you can. The only semidirect product $S_3 \rtimes C_2$ is $S_3 \times C_2$. $\endgroup$ – Derek Holt Mar 23 '16 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.