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I've been watching a video proving $S_4 \simeq$ Aut$(A_4)$, but I was having trouble making sense of it. I wasn't quite sure I even understood exactly what I didn't understand, so I set out to prove it on my own. If I have the core concept down, I'm still unhappy about having to rely on $C_{S_4}(A_4) = \lbrace e \rbrace$ (the centralizer of $A_4$ in $S_4$). It seemed inutitive (which is nice), but I was hoping to rely on something a little more readliy available. Is there a way to easily see the bijection part?

To prove $S_4 \simeq$ Aut$(A_4)$ we need a bijective homomorphism $\alpha: S_4 \rightarrow$ Aut$(A_4)$. Let $\sigma \in S_4$ and define $\alpha(\sigma)=C_\sigma$ where $C_\sigma$ is an autmorphism of $A_4$ defined by $C_\sigma = \sigma h \sigma^{-1}$, $\forall h \in A_4$. We see $C_\sigma$ is an automorphism of $A_4$, since $A_4 \triangleleft S_4$.

First, to show $\alpha$ is a homomorphism: let $\sigma_1 \sigma_2 \in S_4$. So, $\alpha(\sigma_1 \sigma_2) = C_{\sigma_1 \sigma_2}$. Then we have, $\forall h \in A_4$, $C_{\sigma_1 \sigma_2}(h)=$

$$\sigma_1 \sigma_2 h (\sigma_{1} \sigma_{2})^{-1}$$ $$= \sigma_1 \sigma_2 h \sigma_{2}^{-1} \sigma_{1}^{-1}$$ $$= \sigma_1 C_{\sigma_{2}}(h) \sigma_{1}^{-1}$$ $$= C_{\sigma_{1}}[C_{\sigma_{2}}(h)]$$ $$= C_{\sigma_{1}} \circ C_{\sigma_{2}} (h) \;.$$

We can now see $\forall h \in A_4$, $C_{\sigma_1 \sigma_2} = C_{\sigma_{1}} C_{\sigma_{2}}$ so that $\alpha(\sigma_1 \sigma_2) = C_{\sigma_1 \sigma_2} = C_{\sigma_{1}} C_{\sigma_{2}} = \alpha(\sigma_1) \alpha(\sigma_2)$, and $\alpha$ is a homomorphism.

To see $\alpha$ is bijective, we first note that $\mid$Aut$(A_4)$$\mid \; = a \le 24$ since the generators of $A_4 = \; \lbrace(1 2 3), (1 2)(3 4)\rbrace$ give a total of $8$ choices for the $3$-cycle and $3$ choices for the disjoint transpositions (since generators must preserve order under autmorphisms). As $\mid \;S_4 \mid \; = 24$, if we can show $\alpha$ is one-to-one, then $\alpha$ is a bijection.

To this end, let $k \in$ ker$(\alpha)$. Then $\alpha(k) = C_k$, and $\forall h \in A_4$

$$C_k (h) = khk^{-1} = h$$ $$\implies$$ $$kh=hk\;.$$

But, $C_{S_4}(A_4) = \lbrace e \rbrace$ (the centralizer of $A_4$ in $S_4$), (I got desperate here, so I looked this up), so $k=e$. Since ker$(\alpha)$ is trivial, $\alpha$ is one-to-one, and thus a bijection. From earlier, we also know $\alpha$ is a homomorphism, and so $S_4 \simeq$ Aut$(A_4)$.

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  • $\begingroup$ Some general things to keep in mind. 1. We don't need a bijective homomorphism $\alpha : S_4\to \mbox{Aut}(A_4)$, but it certainly suffices to construct one such mapping. 2. For ease of readability, when you define $\alpha$, the image $\alpha (\sigma)$ is a mapping. Define it such that you show to the reader what this mapping does to an element of $A_4$ i.e $\alpha (\sigma) (a) := C_\sigma (a)$ or however it is supposed to work. Is $C_G(A_4)$ the center of $A_4$? $\endgroup$ – Alvin Lepik Nov 28 '17 at 5:45
  • $\begingroup$ Thanks for the advice. I was using $C_G(A_4)$ as the centralizer of $A_4$ in $S_4$, or all the elements in $S_4$ that commute with every element in $A_4$. I meant, $C_{S_4}(A_4)$ $\endgroup$ – user4396386 Nov 28 '17 at 5:48
  • $\begingroup$ One of the comments to the question here: math.stackexchange.com/questions/820704/c-s-4a-4-1?rq=1 has a concise proof of why $C_{S_4}(A_4) = \lbrace e \rbrace$ $\endgroup$ – user4396386 Nov 28 '17 at 6:03
  • $\begingroup$ Instead of showing that $\alpha$ is one-to-one why not show it is onto? $\endgroup$ – Stephen Meskin Nov 29 '17 at 6:41
  • $\begingroup$ Not positive about this, but I think it wouldn't work: $\mid$ Aut$(A_4)$ $\mid $$\; = a \le 24$ and $\alpha$ onto doesn't rule out the possibility of two or more of the automorphisms being the same. It's not true in this case, but if we didn't know exactly what $\sigma$ did, it could be sending them all to the identity automorphism. $a \le 24$ would still be true, and $\alpha$ could be onto. Then Aut$(A_4)$ could be trivial or who knows. $\endgroup$ – user4396386 Nov 29 '17 at 7:52

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