Uniform pro-p groups as a semi-direct product

Question

Let $G$ a finitely generated uniform pro-$p$ group. Then $G/[G,G]$ is abelian and so it is of the form $\mathbb{Z}_p^r\times T$ for some integer $r$ and finite $p$-group $T$. Therefore, $[G,G]$ is contained inside a normal subgroup $H$, with finite index, such that $G/H\cong\mathbb{Z}_p^r$. My question is, under what circumstances is $G$ the semidirect product of $H$ and $\mathbb{Z}_p^r$? I think this is true for $r=1$. Are there other more general conditions that would permit such an isomorphism?

Thanks!

Answer 1

I'm not sure what kind of conditions you have in mind but for every $r>1$ there is a uniform pro-$p$ group $G$ that is not a semidirect product of $H$ and $\mathbb{Z}_p^r$.

To see this consider the uniform pro-$p$ group of dimension $r+1$ with generators $x,y,z,a_1,\ldots,a_{r-2}$ (for some $r\geq 2$) with $z,a_1,\ldots,a_{r-2}$ all central and $xyx^{-1}y^{-1}=z^p$. Then $G'$ is generated by $z^p$ and so $H$ is generated by $z$ and $G/H\cong \mathbb{Z}_p^{r}$. However it is easy to see that any subgroup of $G$ isomorphic to $\mathbb{Z}_p^r$ contains some non-trivial power of $z$.

Indeed, I think a similar argument works for any nilpotent $G$ that is not abelian.