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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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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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$S_4$--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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    Yes, indeed. See my edit.2012-12-11
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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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    @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 $n$.2012-12-11