I can show Inner Automorphism of $S_3$ are isomorphic to $S_3$ .Why every automorphism of $S_3$ is inner automorphism only? That is why there is no automorphism of $S_3$ exist that is not inner automorphism .
Why this is not in case of $S_6$?
Thanks in advanced.
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Curious student
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I know that but why all automorphism of $S_3$ are inner only That was my question.Can you help me please? – Curious student May 02 '18 at 10:41
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1Any automorphism of $S_3$ must permute its three elements of order $2$. Since they generate $S_3$ this means that there are at most $6$ automorphisms. But there are $6$ inner automorphisms. – Derek Holt May 02 '18 at 11:05
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2See here for the case $n \ne 6$. – Derek Holt May 02 '18 at 11:07
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In $S_3$, every automorphism must fix all the conjugacy classes, because distinct conjugacy classes correspond to distinct elements' orders (automorphisms preserve elements' order). So, each automorphism is class-preserving, hence cycle type-preserving. In particular, it sends transpositions to transpositions, and hence it is inner.
In $S_4$ and $S_5$, the classes of the transpositions and the double transpositions have different sizes. So, again, every automorphism is transpositions-preserving (automorphisms preserve classes' size), hence it is inner.
On the other hand, $S_6$ fulfils the necessary conditions for a non-inner automorphism to exist, namely the existence of conjugacy classes of elements of order $2$, which have also the same size: the classes of transpositions and triple transpositions both have size $15$. This doesn't prove yet that such non-inner automorphisms exist (swapping the two classes), but at least makes plausible their existence, here.
citadel
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1Now the claim is that "cycle type preserving" (which for $S_n$ is just another way of saying conjugacy class preserving) is equivalent to inner, but again, that's not necessarily obvious. In $A_3$ all automorphisms are cycle-type preserving, but not all of them are inner. As I recall, proving this takes a bit more work: you usually prove that an automorphism that sends transpositions to transpositions must be inner (and it is not obvious). – Arturo Magidin Nov 16 '23 at 01:50
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1You can take care of $n\lt 6$ and $n\gt 6$ together using that fact: transpositions must be sent to a product of an odd number of disjoint transpositions (for order-preservation, and so that not every element gets mapped to an even permutation). For $n\lt 6$, this forces transpositions to be mapped to transpositions. For $n\geq 6$, class preservation means the set of all transpositions would have to be mapped to the set of all products of $2k+1$ transpositions for a fixed $k$, and size considerations means that the only non-inner possibility is $k=1$, $n=6$. – Arturo Magidin Nov 16 '23 at 02:53