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I have to prove that $S_{4}$ is not isomorphic to $C_{2}\times A_{4}$ and I have no idea to do it. Some ideas?

I have tried a lot of methods but with no luck.

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    There are so many ways in which the two groups differ that an important question is "how far do you want to go?" You could loo$k$ at orders of elements and see that the set of possible orders is different; you could loo$k$ at orders of subgroups and see the set of possible sizes is different; you could loo$k$ at im$p$ortant subgroups (the center, the commutator subgroup) or quotients (the abelianization) and see that they are different; etc.2012-03-06

3 Answers 3

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$C_2\times A_4$ has an element of order 6. $S_4$ doesn't.

(On the other hand, $S_4$ has an element of order 4, which $C_2\times A_4$ doesn't).

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    Thank you for the help and the great observation.2012-03-06
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Check that $S_4$ does not admit a surjective homomorphism onto $A_4$, because $S_4$ has no normal subgroup of order $2$.

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If $S_4$ were isomorphic to $C_2\times A_4$ via $f\colon C_2\times A_4\to S_4$, then the image of $(1,0)$ would be an element of order $2$ that is central (commutes with everything) in $S_4$.

In $S_4$, there are two types of elements of order $2$: transpositions, and products of two disjoint transpositions.

If $i,j\in\{1,2,3,4\}$, $i\lt j$, does $(ij)$ commute with everything in $S_4$?

If $(ij)(k\ell)\in S_4$ is a product of two disjoint transpositions, does it commute with everything in $S_4$?