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I already proved this, but I think I can reduce my solution.

My solution : There are 4 Sylow 3-subgroup of $S_4$, and denote the set of Syl 3-subgroups by $P=\{P_1,P_2,P_3,P_4\}$.

Then, by a group action $\operatorname{Aut}(S_4) \times P \to P$ defined by $(f,P_i) \to f(P_i)$, obtain a homomorphism $\phi:\operatorname{Aut}(S_4)\to S_4(=\operatorname{Perm}(P))$.

To show that $\ker(\phi)=0$, I suppose $f \in \operatorname{Aut}(S_4)$ fix every Sylow 3-subgroup.

Then I can derive that $f$ fixes every 3-cycle rather easily.

But proving that $f$ fixes $2$-cycle is very long and not seems good, and my question arises here. (I consider every case that $f(ab)$ can be)

Does anyone have a smart idea for proving $f$ fixes 2-cycle?

Shaun
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Yongha
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    I know this isn't what you specifically want, but isn't there a slick way of proving this by just defining the isomorphism? $ S_4 \rightarrow \text{Aut}(S_4) $; I may be wrong but does $ \sigma \rightarrow \phi_\sigma $ where $\phi_\sigma(\tau) = \tau \sigma $ do the trick? – JC574 Jul 28 '14 at 18:18
  • If $f$ fixes every $3$-cycle, then in particular $f$ fixes $A_4$ pointwise, since $3$-cycles generate $A_4$. Then just prove that $f$ fixes some odd permutation and you are done. – spin Jul 28 '14 at 18:22
  • Maybe something could be done by decomposing $(abc)$ into $(ac)(bc)$. – Alexander Gruber Jul 28 '14 at 18:26
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    @JC574 That is not an automorphism. – Tobias Kildetoft Jul 28 '14 at 18:26
  • ha no it isn't! that was silly – JC574 Jul 28 '14 at 18:29
  • How many cases did you consider? Should there be just $f(ab)=(ab)$, $f(ab)=(ac)$ and $f(ab)=(cd)$? – Quang Hoang Jul 28 '14 at 18:29
  • @JC574 But note that you were actually quite close. You needed $\phi_{\sigma}(\tau) = \sigma\tau\sigma^{-1}$. – Tobias Kildetoft Jul 28 '14 at 18:30
  • :) yeah i just wrote that! or rather $\sigma^{-1} \tau \sigma $ – JC574 Jul 28 '14 at 18:30
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    @JC574: That gives you an isomorphism, but you need proof that there are no outer automorphisms of $S_4$. This is not obvious, for example $S_6$ has an outer automorphism – spin Jul 28 '14 at 18:32
  • A quick google search shows: http://math.stackexchange.com/questions/8917/automorphism-of-s4 – Quang Hoang Jul 28 '14 at 18:32
  • can you explain to me a bit why we need this extra proof of no outer automorphisms if we have the isomorphism? – JC574 Jul 28 '14 at 18:34
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    @JC574: The map $\sigma \mapsto \phi_\sigma$ is injective. You should still prove that it is surjective to conclude that it is an isomorphism. – spin Jul 28 '14 at 18:35
  • Ah obviously. for some reason in my head i thought we had already $| \text{Aut}(S_4) |=|S_4| $ – JC574 Jul 28 '14 at 18:37
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    @JC574 with the ${}^{-1}$ on the first of the terms, the map $\sigma\mapsto \phi_{\sigma}$ becomes an antihomomorphism rather than a homomorphism. – Tobias Kildetoft Jul 28 '14 at 18:38

3 Answers3

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Assuming that you know that $f$ fixes every $3$-cycle, it follows that $f$ also fixes every even permutation (because $A_4$ is generated by $3$-cycles).

Now $A_n$ is the commutator subgroup of $S_n$ when $n \geq 3$ (proof: write a $3$-cycle as the commutator of two transpositions). So the parity of an permutation is invariant under an automorphism of $S_n$.

Thus $f$ permutes the transpositions $(12), (13), (14), (23), (24), (34)$.

Now $(123) = f(123) = f((13)(12)) = f(13)f(12)$ so $f(12)$ is $(12)$ or $(13)$ or $(23)$.

Also $(12)(34) = f((12)(34)) = f(12)f(34)$ so $f(12)$ is $(12)$ or $(34)$.

Thus $f(12)$ is $(12)$ hence $f = 1$.

spin
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  • You may do the problem by knowing that every automorphism of $S_4$ is inner and of course we have $Z(S_4)={1}$. Now, if you use the straightforward consequence of N/C Lemma saying that:

    $Inn(G)\lhd G$ and $G/Z(G)\cong Inn(G)$

    then you'll find that $Aut(S_4)\cong S_4$.

     – Mikasa Jul 28 '14 at 21:09
  • How can "$A_n$ is the commutator subgroup of $S_n$" imply "the parity of an permutation is invariant"? I can't understand that. – Yongha Jul 29 '14 at 01:19
  • @Yongha: Sorry, the point is that $A_n$ is characteristic in $S_n$. In other words if $f$ is any automorphism of $S_n$, then for all $\sigma \in S_n$ we have $f(\sigma)$ even if and only if $\sigma$ even – spin Jul 29 '14 at 06:52
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Let's denote a conjugacy class of representative $\sigma\in S_4$ as: $$\operatorname{cl}(\sigma,o(\sigma),\text{cycle type},\text{size})$$ The conjugacy classes of $S_4$ are then: \begin{alignat}{1} &\operatorname{cl}((),1,(1,1,1,1),1) \\ &\operatorname{cl}((12),2,(2,1,1),6) \\ &\operatorname{cl}((12)(34),2,(2,2),3) \\ &\operatorname{cl}((123),3,(3,1),8) \\ &\operatorname{cl}((1234),4,(4),6) \\ \tag1 \end{alignat} Since automorphisms send conjugacy classes to conjugacy classes, while preserving elements' order and classes' size, from $(1)$ follows that every element of $\operatorname{Aut}(S_4)$ is class-preserving, hence (again by $(1)$) cycle type-preserving, and hence inner. Therefore: $$\operatorname{Aut}(S_4)=\operatorname{Inn}(S_4)\stackrel{Z(S_4)=1}{\cong}S_4$$

This argument firstly (uniquely?) fails for $S_6$, where: \begin{alignat}{1} &\operatorname{cl}((12),2,(2,1,1,1,1),15) \\ &\operatorname{cl}((12)(34)(56),2,(2,2,2),15) \\ \end{alignat} and hence, in principle, some non-inner automorphism might swap the two classes.

Kan't
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You can also choose a good representation of $S_4$ as $<(12),(1234)>$ to restrict the size of $\text{Aut}(S_4)$. Since $S4\cong\text{Inn}(S_4)\leqslant\text{Aut}(S_4)$, to obtain the result, it suffices to have $|\text{Aut}(S_4)|\leqslant24$.

Generators are mapped to elements of the same order under the isomorphism.So let us consider elements of order $2$ and order $4$.

The element $(12)$ is mapped to an element of order $2$ under the isomorphism, which must be of the form $(..)$ or $(..)(..)$. Since automorphisms send conjugacy classes to conjugacy classes of the same length, elements of the form $(..)(..)$ is ruled out.

Since $(12)(1234)$ has order $3$, the elements $(13)$ and $(24)$ are also ruled out, because $(13)(1234)$ and $(24)(1234)$ have order $2$.

In summary, $(12)$ and $(1234)$ have $4$ and $6$ choices, respectively. This implies that $|\text{Aut}(S_4)|$ has $24$ elements at most. Comparing the orders of $\text{Aut}(S_4)$ and $S_4$ yields the answer.

mum cuc
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