# Nonabelian group of order $p^3$ and semidirect products

Let G be a nonabelian group of order $p^3$ where p is an odd prime. Suppose that G contains an element of order $p^2$. Then G is isomorphic to the semidirect product $Z_{p^2} \rtimes_{\alpha} Z_p$, hwere $\alpha: Z_p \rightarrow Z^x_{p^2}$ is induced by $\alpha(\bar{1})=\overline{p+1}$.

Proof Let b ∈ G have order $p^2$. Then has index p in G, so is normal in G by Proposition 4.1.12. Thus, G is an extension of $Z_{p^2}$ by $Z_p$, via $f : G → Z_p$. Since G is not abelian, Lemma 4.7.15 shows that the induced homomorphism α : $Z_p$ → Z × $p^2$ = $Aut(Z_p^2)$ is nontrivial....

Proposition 4.1.12. Let G be a finite group and let p be the smallest prime dividing the order of G. Then any subgroup of G of index p is normal.

Lemma 4.7.15 Let f : G → $Z_k$ be a central extension of the abelian group H by the cyclic group $Z_k$. Then G is abelian.

I'm a bit confused about this proof. I'm not sure how the fact that is normal in G implies that G is an extension. I understand that, in order for G to be an extension, must be normal (and 4.1.12 obviously shows that)...but how is this sufficient by itself to conclude that G is an extension? Don't we also need to prove that $f: G \rightarrow Z_p$ is a homomorphism?

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Typo? If we have $o(x)=p^3=|G|$ then isn't $G$ cyclic? – user47805 Mar 22 '13 at 10:11
@user47805 Sorry. There was a typo on the second line. It was supposed to say that G has an element of order $p^2$ instead of $p^3$. – user58289 Mar 22 '13 at 10:36
related – Alexander Gruber Mar 22 '13 at 11:13
@Artus, I have fixed that typo for you. – Andreas Caranti Mar 22 '13 at 11:15

This is just an addition to Andreas Caranti's answer.

I don't know how much you know about semidirect products and exact sequences, but as I like this view on semidirect products the most, we will try it:

A group $G=H\rtimes K$ is the semidirect product of two groups $H$ and $K$ iff there is a exact sequence of groups $$0\rightarrow H\rightarrow G\rightarrow K\rightarrow 0,$$ that splits.

Now we see what happens: $Z_{p^2}$ is a normal subgroup, so there is an exact sequence $$0\rightarrow Z_{p^2}\rightarrow G\rightarrow G/(Z_{p^2})\rightarrow 0$$ and as $Z_{p^2}$ has index $p$ in $G$, we get that $G/(Z_{p^2})$ has $p$ elements, so it's just isomorphic to $Z_{p}$.

So what is left? This sequence has to split and it splits iff there is an element of order $p$ in $G\setminus H$. This is exactly, what Andreas Caranti showed.

Edit: Artus asked, what Lemma 4.7.15 has to do with the fact, that the induced homomorphism $\alpha$ is not trivial. I will do this in a more general context. If $G\cong H\rtimes_{\alpha} K$, then $H$ is a normal subgroup in $G$, $K$ is just a regular subgroup in $G$ and $\alpha(k)$ is just the conjugation with $k$ in $G$ restricted to $H$, which is an element of $Aut(H)$ as $H$ is normal. So what about your situation? $\alpha$ is then trivial, iff the conjugation with all elements of $K$ restricted to $H$ is trivial, what is the case iff $H$ commutes with $K$ and as $H$ is abelian in your case, $H$ commutes with $H$ too, so $H$ lies in the center of $G$. Lemma 4.7.15 says, that this is not possible, so $\alpha$ is not trivial.

I jused some identifications like $H\cong H\times\{1\}$ and so on...

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Precisely, thanks. Just edit the ending, "This sequence has to split and it splits iff there is an element of order $p$ in $G \setminus H$." – Andreas Caranti Mar 22 '13 at 11:37
@archipelago I just have one more question. When the proof says "Since G is not abelian, Lemma 4.7.15 shows that the induced homomorphism α : $Z_p$ → Z × $p^2$ = $Aut(Z_p^2)$ is nontrivial". I'm not really sure how that's related to Lemma 4.7.15...because all that does is to conclude that G is abelian, which is not true in this case. So how does the fact that "G is not abelian" allow us to use Lemma 4.7.15? – user58289 Mar 22 '13 at 11:43
Artus, I edited my answer. – archipelago Mar 22 '13 at 12:22

One thing that appears to be missing is the proof that there is an element of order $p$ in $G \setminus H$, where $H = \langle b \rangle$.

In fact if $H$ is a normal subgroup of the group $G$, then by definition $G$ is an extension of $H$ by $G/H$. In this particular case you want to show that the extension splits, that is, there is a complement to $H$ in $G$.

Take any element $a \in G \setminus H$.

First note that $\langle b^{p} \rangle$ is a normal subgroup of $G$ (since it is characteristic in $H$, which is normal in $G$), with a quotient of order $p^{2}$, which is thus abelian. Since $G$ is non-abelian, it follows that $G' = \langle [a, b] \rangle = \langle b^{p} \rangle$.

If it has order $p$, we are done. If it has order $p^{2}$, then $a^{p}$ is an element of order $p$ of $H$, so $a^{p} = b^{ip}$ for some $i \not\equiv 0 \pmod{p}$. Now compute $$(a b^{-i})^{p} = a^{p} b^{-ip} [b, a]^{- i \binom{p}{2}} = 1,$$ so $a b^{-1} \in G \setminus H$ has order $p$. Here we are using the fact that $p$ divides $\binom{p}{2}$, as $p$ is odd. (The argument fails in fact in the quaternion group of order $8$.)

Of course $a$ cannot have order $8$, otherwise $G$ would be cyclic.

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Thanks a lot. I thought we had to check two conditions (that H had to be normal and the result had to be a homomorphism) to show that it was an extension...I didn't know that H being normal automatically implied that there was a homomorphism and thus an extension. – user58289 Mar 22 '13 at 11:37