# The groups $S_n$ and $A_n \times \mathbb{Z}_2$ are not isomorphic for $n>2$

How can I prove the following statement:

$S_n$ and $A_n \times \mathbb{Z}_2$ are not isomorphic for every $n>2$.

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Since you are new to this site, please consider reading this: How to ask a homework question? I wrote this comment because the question sounds homework-like. –  Julian Kuelshammer Oct 16 '12 at 18:22
Thank you very much! –  Doeshka Oct 20 '12 at 21:13

A simple idea is to count the number of elements with order $2$ in the two groups. $g\in S_n$ has order $2$ iff it is product of disjoint transpositions, $h=(j,k)\in A_n\times\mathbb{Z}_2$ has order $2$ if $j$ is a product of an even number of disjoint transpositions, so, if $n\geq 4$, the elements of order $2$ in $S_n$ are striclty more than the elements of order $2$ in $A_n\times\mathbb{Z}_2$ and the two groups cannot be isomorphic. For $n=3$ we have that $A_3\times \mathbb{Z}_2$ is isomorphic to the abelian group $\mathbb{Z}_6$ while $S_3$ is not an abelian group.

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Like this way(+1) –  Babak S. Oct 15 '12 at 15:29

If you take $Q=\langle (1,2)\rangle\subset S_n$ then it can be a complement of $A_n$ in $S_n$. So $S_n$ is a semidirect product of $A_n$ by $Q$.

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$A_n \times \mathbb{Z}_2$ is a semidirect product too. –  user17794 Oct 15 '12 at 15:17
@TimDuff: $Q\cong\mathbb Z_2$. ;-) –  Babak S. Oct 15 '12 at 15:22

One way to do this is to show that $S_n$ does not have a normal subgroup of size 2 (since the direct product has such a normal subgroup).

The non-trivial element in such a subgroup would be a product of disjoint transpositions. But conjugating such a thing by any transposition which doesn't occur in the permutation but which does contain a digit that occurs in the permutation will change it, and we can ensure such a transposition exists because $n > 2$.

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Another way to see that $S_n$ and $A_n\times \mathbb{Z}/2$ are not isomorphic is that the latter contains an element of order $2$ which commutes with every other element. However, if $n>2$ the only element of $S_n$ that commutes with all other elements is the identity. (To see this, write down the cycle decomposition of your candidate permutation, and pick a transposition that either intersects two cycles or fits inside a cycle of length at least $3$.)

In other words, $A_n\times \mathbb{Z}/2$ has a nontrivial center, while the center of $S_n$ is trivial.

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+1 This is probably the easiest, most direct proof using the fact that $\,S_n\,$ is centerless. –  DonAntonio Oct 16 '12 at 3:00