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in my notations, $S_n$ is the symmetric group and $A_{n+1}$ the alternating group.

I showed that if $n$ is even it is never possible because you need to have $n! | n! \cdot\dfrac{n+1}{2}$.

So we need $n$ odd. I also showed that it is impossible for $n=3$ because $A_4$ has no subgroups of order 6.

Now, for other odd values of $n\ge 5$, we can use the fact that $A_{n+1}$ is simple, but how to go on from here ?

The links given as duplicate do not provide a clear answer and I am still stuck on this. Here is what I have :

If you consider transposition $(1\quad 2)$, the permutations that commute with this transposition are the ones that fix 1 and 2, and the ones that switch 1 and 2. So that the size of the centralizer of $(1\quad 2)$ is $2(n-2)!$.

Now, because $(1\quad 2)$ is of order 2, it's image must also be an element of order 2 in $A_{n+1}$, so that it must be the product of $2k$ disjoint transpositions. If we now consider the centralizer of the product of $2k$ disjoint transpositions in $A_{n+1}$, we find that a permutation in the centralizer must fix or switch each of the transpositions, leaving $2^{2k}$ choices. Then, we must either pick a permutation of $A_{n+1-2k}$ or $S_{n+1-2k}\setminus A_{n+1-2k}$ to make sure the parity is even. Leaving in total : $$ (n+1-2k)!\cdot 2^{2k-1}$$ Permutations in the centralizer of the product of $2k$ transpositions.

If we had $(n+1-2k)!\cdot 2^{2k-1}< 2(n-1)!$ we would be done, but it is not always true.

aleph0
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  • The link given as duplicate does not provide an answer to my question, why did my question get closed ? – aleph0 Mar 31 '19 at 19:29
  • Derek Holt answers the question at this duplicate, too. In the first duplicate, the answer is the Proposition, in Arturo's answer. So there is a solution given. – Dietrich Burde Mar 31 '19 at 19:29
  • In arthuro's answer, he doesn't prove his proposition, and only states that my statement is false for $n=1$. I tried reading Derek's answer, and I'm not convinced by the fact that $2(n-2)!$ is greater than the centralizer of any permutation of order 2. A permutation of order 2 in $A_{n+1}$ is the product of $2k$ disjoint transpositions. So that the size of the centralizer of a permutation is $(n+1-2k)!\cdot 2^{2k-1}$ if I am not mistaken. If you take $k =1$, then $2(n-2)! \le 2(n-1)! $. Maybe my formula for the centralizer of a product of 2k disjoint transposions in $A_{n+1}$ is wrong ? – aleph0 Mar 31 '19 at 19:58
  • Yes it's wrong! It should be $(n+1-4k)!\cdot (2k)!\cdot 2^{2k-1}$. – Derek Holt Apr 01 '19 at 07:10

1 Answers1

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If we had $S_n$ isomorphic to a subgroup of $A_{n+1}$, then it would have index $\frac{n+1}{2}$, so the action of $A_{n+1}$ on the cosets of this subgroup gives a nontrivial group homomorphism $A_{n+1}\rightarrow S_{\frac{n+1}{2}}$, which is not possible for $n\geq 4$, since $|A_{n+1}|>|S_{\frac{n+1}{2}}|$, and $A_{n+1}$ is simple.

Chris H
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