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For what $n$ is $A_n$ a marginal subgroup of $S_n$? Strict definition of marginal subgroups and brief overview of their properties can be found here

What have I tried:

$A_n$ is characteristic in $S_n$ for all $n$ as it is its only subgroup of index $2$. Thus the fact that each marginal subgroup is characteristic will not help us to screen out anything.

Thus, we may probably consider some stronger properties, such as, for example, direct power closed characteristicity... However, I failed to achieve any success in that direction.

Also, for $n = 2$, $A_n$ is the trivial subgroup and thus indeed marginal. However, that does not tell us anything about $n > 2$...

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1 Answer 1

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For $n>2$, $A_n$ is not marginal.

Suppose it were, for a word $w$. Any two conjugate elements $x,x'\in S_n$ have the same parity, so there are elements $a,b\in A_n$ such that $x'=ax=xb$. So $w(x_1,\dots,x_n)$ is fixed by conjugation (of $x_1,\dots,x_n$, and hence of $w(x_1,\dots,w_n)$) by any element of $S_n$. Hence $w$ can only take values in the centre of $S_n$, which is trivial for $n>2$.

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