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?
Thanks in advance
Best Answer
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...