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I do know of $3$ classes of groups (up to isomorphism) of order $24$ that are commutative (direct products of $\mathbb{Z}$/(factors of $24$)$\mathbb{Z}$. Can you just take the semi direct product instead of the direct product and make these groups non commutative?? Does this normally guarantee non-commutative groups?

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3 Answers 3

up vote 2 down vote accepted

Taking your questions in order:

Yes, there exist noncommutative groups of order $24$: $S_4$ for example.

No, of course the fact that you know some commutative groups of order $n$ does not prove that every group of order $n$ is commutative: why would it?

Yes, a non-trivial semidirect product is always noncommutative. However, non-trivial products do not always exist. For example, there are none for $\Bbb{Z}/3\Bbb{Z}$ and $\Bbb{Z}/8\Bbb{Z}$ with $\Bbb{Z}/8\Bbb{Z}$ normal, but there are with $\Bbb{Z}/3\Bbb{Z}$ normal.

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In this case, I forgot the clear counterexample of $S_4$, as all of you have mentioned, but there are certain orders, where groups can be either commutative or non commutative, but not both. –  chubbycantorset Dec 11 '12 at 21:00
    
@chubbycantorset: What? A group is always commutative or noncommutative, but not both. If you mean every group of a given order $n$, then note that it's impossible for every group of order $n$ to be noncommutative (cyclic groups exist). It is possible for every group of order $n$ to be commutative (e.g. when $n$ is a prime), but this certainly isn't implied by the fact that there are some commutative groups of order $n$. –  Chris Eagle Dec 11 '12 at 21:04

$S_4$--the group of permutations on $4$ elements--is a non-commutative group of order $24$.

Another example is the dihedral group of order $24$, which can be obtained as a semidirect product of the cyclic groups of order $2$ and order $12$. (Recall that cyclic groups are abelian.)

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Ah, right. Of course. But can you construct non commutative groups from taking the semi direct products, instead of taking the direct product in the groups that I mentioned? –  chubbycantorset Dec 11 '12 at 20:52
    
Yes, indeed. See my edit. –  Cameron Buie Dec 11 '12 at 20:56

Oh, lots of them. In fact, up to isomorphism, there are $\,12\,$ different non-abelian groups of order $\,24\,$ , and perhaps the easiest one is $\,S_4\,$ , or the dihedral $\,D_{12}\,$ ...

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