Let $G$ be a non-cyclic group of order $p^3$ for an odd prime $p$. Prove that $G \simeq H \rtimes_{\theta}K$, where $H$ is a normal subgroup of $G$ of order $p^2$, $K$ is a subgroup of order $p$, and $\theta : K \to Aut(H)$ is a homomorphism.

I managed to prove that there exists a normal subgroup $H$ of order $p^2$. Then I took some $g \in G-H$. If $g$ is of order $p$, I am done since $G \simeq H \rtimes \langle g \rangle $. But what if all $g \in G - H$ are of order $p^2 $?

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    $\begingroup$ Impossible that all the non-trivial elements in $\;G\;$ are of order $\;p^2\;$ since there exists a subgroup of order $\;p\;$ ... $\endgroup$ – DonAntonio Dec 23 '18 at 22:10
  • $\begingroup$ Sorry, I meant all $g \in G - H$. I edited my question $\endgroup$ – user401516 Dec 23 '18 at 22:18
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    $\begingroup$ Observe that the conclusion would be false when $p=2$. The quaternion group $Q_8$. In other words, the fact that $p$ is odd must come into play somehow. $\endgroup$ – Jyrki Lahtonen Dec 24 '18 at 6:36
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    $\begingroup$ The claim seems to follow from the observations made by the OP and the argument given in this old answer by Arturo Magidin. $\endgroup$ – Jyrki Lahtonen Dec 24 '18 at 7:21

The result follows immediately if all elements of $G$ have order $1$ or $p$, so we can assume that there exists an element of order $p^2$. This generates a cyclic subgroup of index $p$, which is normal in $G$, so we can assume that $H$ is cyclic of order $p^2$.

So let $g \in G \setminus H$, and assume that $g$ has order $p^2$ (if it has order $p$ then we are done). So $g^p = h^p$ for some $h \in H$. We claim that $(gh^{-1})^p=1$, and hence $gh^{-1}$ has order $p$ and we are done.

This is immediate if $G$ is abelian, so suppose not. Then $[G,G]=Z(G)$ has order $p$, so commutators are central, and $(xy^{-1})^p = x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ (where $[a,b]$ denotes the commutator $a^{-1}b^{-1}ab$).

Then, since $p$ is odd and $[y^{-1},x]$ has order $p$, we get $[y^{-1},x]^{p(p-1)/2}=1$ and the claim follows. (Note that this result is false when $p=2$, and the quaternion group $Q_8$ is a counterexample.)

  • $\begingroup$ Thank you for your answer. Can you please explain why $(xy^{-1})^p=x^py^{-p}[y^{-1},x]^{p(p-1)/2}$ ? $\endgroup$ – user401516 Dec 24 '18 at 12:30
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    $\begingroup$ If $[G,G] \le Z(G)$ then $(ab)^k = a^kb^k[b,a]^{k(k-1)/2}$ for all $a,b \in G$ and all $k \ge 0$. Prove it by induction on $k$, using $ba = ab[b,a]$. $\endgroup$ – Derek Holt Dec 24 '18 at 15:24
  • $\begingroup$ Where did you use the fact that $H$ is cyclic? $\endgroup$ – user401516 Dec 24 '18 at 21:45
  • $\begingroup$ The existence of $h \in H$ such that $g^p=h^p$. $\endgroup$ – Derek Holt Dec 24 '18 at 22:45

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