$\mathbb Z_{p^2}$ is not a non trivial semidirect product.

Question

I am trying to prove that the group $\mathbb Z_{p^2}$ (p prime) is not a non trivial semidirect product. Since a group $G \cong K \rtimes H$ if and only if for all short exact sequences $$0 \rightarrow H \rightarrow G \rightarrow K \rightarrow 0$$splits on the right, then I thought that I could prove that the appropiate sequence doesn't split on the right.

The only non trivial group of $\mathbb Z_{p^2}$ is $\mathbb Z_p$, so the only possible non trivial semidirect product is $\mathbb Z_p \rtimes \mathbb Z_p$. I define the sequence $$0\longrightarrow \mathbb Z_{p} \overset{f}{\longrightarrow} \mathbb Z_{p^2} \overset{g}{\longrightarrow} \mathbb Z_p \longrightarrow 0 $$ Where $f$ is the inclusion an $g$ is the "projection".

I would like to show that there is no such $\beta: \mathbb Z_p \to \mathbb Z_{p^2}$ with $g \circ \beta =Id_{\mathbb Z_{p}}$.

Any help with this part of the solution would be appreciated.

Answer 1

$\mathbb{Z}_{p^2}$ has a unique subgroup of order $p$, and that subgroup is the kernel of the homomorphism $g$.

Answer 2

I think you meant $\;G:=\Bbb Z_{p^2}\;$ is the cyclic group of order $\;p^2\;$ , and thus any "semi" direct product would in fact be direct, so direct sequences, split or not, are way too heavy and unnecessary for this.

Since the only non-trivial subgroup of $\;G\;$ is the cyclic one of order $\;p\;$, any non-trivial direct product will have exponent at most $\;p\;$ and this is impossible.

Answer 3

Note that $\left|\operatorname{Aut}(\mathbb Z_p)\right|=p-1$, so the only homomorphism $\varphi\colon\mathbb Z_p\to\operatorname{Aut}(\mathbb Z_p)$ is the one with $\ker\varphi=\mathbb Z_p$, namely the trivial one (the only other option would be $\ker\varphi=\{0\}$, namely the embedding $\mathbb Z_p\hookrightarrow \operatorname{Aut}(\mathbb Z_p)$, which is prevented by cardinality reasons, being $p>p-1$). Therefore: $$\mathbb Z_p\ltimes\mathbb Z_p=\mathbb Z_p\times\mathbb Z_p\not\cong \mathbb Z_{p^2}$$

(More easily: nontrivial semidirect products are nonabelian, whereas $\mathbb Z_{p^2}$ is abelian. )