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From a little reading, I know that for $p$ and odd prime, there are two nonabelian groups of order $p^3$, namely the semidirect product of $\mathbb{Z}/(p)\times\mathbb{Z}/(p)$ and $\mathbb{Z}/(p)$, and the semidirect product of $\mathbb{Z}/(p^2)$ and $\mathbb{Z}/(p)$.

Is there some obvious reason that these groups are nonabelian?

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    The semidirect product of two groups isn't unique; you need to specify how one group acts as automorphisms on the other. Can you write down two elements of the group that don't commute?2011-12-08
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    a semi-direct product $N\rtimes H$ (when not adirect product) doesnt have $H$ and $N$ commute ($H$ acts by conjugation as automorphisms of $N$)2011-12-08
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    A semidirect product that is not simply a direct product is by definition non-abelian. It is a very specific sort of a non-abelian.2011-12-08
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    A semi-direct product (of $N$ and $H$ with respect to $\phi : H \to {\sf Aut}(N)$) is commutative iff $N,H$ are both commutative and $\phi$ is the identity everywhere.2011-12-08
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    this is as explicit as is gets http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf2011-12-08

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Think about it this way, suppose that you have that $G=A\rtimes_\varphi B$ where $A,B$ are abelian. You then have a short exact sequence $0\to A\to G\xrightarrow{\gamma} B\to 0$ and a backmap $B\xrightarrow{\psi}G$ such that $\gamma\circ\psi=1_B$. If you assume that $G$ is abelian then the splitting lemma for $\mathbb{Z}$-modules tells you that the sequence $0\to A\to G\to B\to0$ splits and so $G\cong A\oplus B$. Thus, if $A\rtimes_\varphi B$ is abelian, then $A\rtimes_\varphi B\cong A\oplus B$. But, it's easy to check that this is the case if and only if $\varphi$ is trivial. So, non-trivial semidirect products induce non-abelian groups.

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    You've got your $\gamma$ over the map $B\rightarrow 0$, which you probably don't mean.2011-12-08
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    Ah, yes, thank you!2011-12-08