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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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  • $\begingroup$ 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? $\endgroup$ – Qiaochu Yuan Dec 8 '11 at 19:01
  • $\begingroup$ a semi-direct product $N\rtimes H$ (when not adirect product) doesnt have $H$ and $N$ commute ($H$ acts by conjugation as automorphisms of $N$) $\endgroup$ – yoyo Dec 8 '11 at 19:03
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    $\begingroup$ 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. $\endgroup$ – Jack Schmidt Dec 8 '11 at 19:03
  • $\begingroup$ 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. $\endgroup$ – Ewan Delanoy Dec 8 '11 at 19:06
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    $\begingroup$ this is as explicit as is gets math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf $\endgroup$ – yoyo Dec 8 '11 at 19:41
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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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    $\begingroup$ You've got your $\gamma$ over the map $B\rightarrow 0$, which you probably don't mean. $\endgroup$ – Thomas Andrews Dec 8 '11 at 20:17
  • $\begingroup$ Ah, yes, thank you! $\endgroup$ – Alex Youcis Dec 8 '11 at 20:19

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