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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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    Find their Sylow 2-subgroups.2012-03-06
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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 look at orders of elements and see that the set of possible orders is different; you could look at orders of subgroups and see the set of possible sizes is different; you could look at important 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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    How can I be sure that $S_{4}$ doesn't have an element of order 6?2012-03-06
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    @bemyguest If $\gamma=\sigma_1\cdots\sigma_n$ where the $\sigma_i$'s are disjoint cycles, then $|\gamma|=\operatorname{lcm}(|\sigma_1|,\ldots,|\sigma_n|)$. This means no element in $S_4$ has order greater than 4.2012-03-06
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    @bemyguest: and if arguments such as SL2's don't convince you, then surely writing down each of the 24 elements of $S_4$ and the order of each one will?2012-03-06
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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 $S_4$?