In short, I am trying to find a faithful action of $N \rtimes G$ on $N$, where I know that the action for the semidirect product is faithful. My first attempt was $(n, g) \cdot n'=(nn') \cdot g$, but I don't think this turns out to be an action--compatibility doesn't seem to work out.

My real goal (please note this is a homework problem) is to show that $S_4 \times C_2 \cong (C_2 \times C_2 \times C_2) \rtimes S_3$, where the action is given by place permutation; that is

$$ (x_1, x_2, x_3) \cdot \alpha=(x_{\alpha^{-1}(1)}, x_{\alpha^{-1}(2)}, x_{\alpha^{-1}(3)}).$$

Most likely there are other ways to approach this problem, but here is what I'm attempting:

I can show that $(C_2 \times C_2 \times C_2) \rtimes S_3 \cong (E \rtimes S_3) \times C_2$, where $E$ is the subgroup of $C_2 \times C_2 \times C_2$ consisting of elements with two nonzero entries, and the identity. If I can find a faithful action of $E \rtimes S_3$ on $E$, then this should induce an isomorphism to $S(E) \cong S(C_2 \times C_2) \cong S_4$, and I believe I'll be done.

To be clear: I'd like to know if in general there is there a faithful action of $N \rtimes G$ on $N$, and if so how to construct it.

Thank you for any help!

  • $\begingroup$ A quick note on notation - you seem to be try to create a left action of $N\rtimes G$ on $N$, but later use a right action of $S_3$ on $C_2\times C_2\times C_2$. It may be less confusing to stick to just right actions $\endgroup$ – Robert Chamberlain Jan 15 '17 at 9:20
  • $\begingroup$ In general there is not such faithful action. This is not possible when $|N|$ is prime, for example, because ${\rm Aut}(C_p)$ is abelian. $\endgroup$ – Derek Holt Jan 15 '17 at 10:28
  • $\begingroup$ The action of $G$ on $N$ in $N\rtimes G$ consists of group automorphisms, but the action of $N\rtimes G$ on $N$ does not need to, it only needs to act on the underlying set of $N$. If we consider for example $D_6\cong C_3\rtimes C_2$ then the action of $D_6$ on $C_3$ can be its natural action on 3 points. $\endgroup$ – Robert Chamberlain Jan 15 '17 at 12:58

Consider the right action of $N\rtimes G$ on the right cosets of $G$ (as a subgroup of $N\rtimes G$) - there are $|N|$ cosets so this induces a homomorphism $N\rtimes G\to S_{|N|}$ as required.

To see this action is faithful let $(n,g)\in N\rtimes G$. $(n,g)=(1,g)(n\cdot g,1)$, so for $n_0\in N$ we have $$\begin{array}{ll} G(n_0,1)(n,g) & =G(n_0,g)(n\cdot g,1)\\ &=G(n_0\cdot g,1)(n\cdot g,1)\\ &=G((n_0n)\cdot g,1) \end{array}$$

If $n\ne 1$ then this give $G(n^{-1},1)(n,g)=G$. If $n=1$, $G$ acts faithfully on $N$ so we can choose $n_0$ with $n_0\cdot g\ne n_0$ so $G(n_0,1)(n,g)=G(n_0\cdot g,1)$. Hence the action is faithful.

Calculating the image in $S_{|N|}$ may be non-trivial in general, but when $|N\rtimes G|=|S_{|N|}|$ we obtain $N\rtimes G\cong S_{|N|}$ as you want.


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