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Let $\mathcal{G}=\mathrm{Gal}(M_{\infty }/F)$. The Galois group sequence isFor $g\in \Gamma$ and $x\in X$, choose a lift $\tilde{g}\in \mathcal{G}$ and defineThis is independent of the lift because $X$ is abelian. Write $X$ additively. As an abelian pro-p group, it is a compact ${\mathbb{Z}}_p$-module, with $ax$ defined by the limit of integer powers, and the conjugation action is continuous and ${\mathbb{Z}}_p$-linear. It extends to the Iwasawa algebraIndeed, use the finite quotients of $X$ obtained by intersecting it with open normal subgroups of $\mathcal{G}$; on each quotient the action factors through one of these finite group rings. This defines the natural compact Galois module structure, even when $X$ is not finitely generated. Choosing $\gamma$ identifies $\Lambda$ with ${\mathbb{Z}}_p[[T]]$ through $T=\gamma -1$.

The subgroup $(\gamma -1)X$ is closed: it is the image of the compact group $X$ under a continuous map. Every element $(\gamma -1)x$ is a commutator of a lift of $\gamma$ with $x$. ConsequentlyConversely, in the quotient by $(\gamma -1)X$, the lift of $\gamma$ centralizes the image of $X$. Its powers are dense in the procyclic subgroup they generate. Since these powers map densely onto $\Gamma$, compactness shows that $\mathcal{G}$ is generated by $X$ and that procyclic subgroup. The quotient is therefore abelian. HenceOne can also see this by choosing a lift of $\gamma$. Since $\mathcal{G}$ is a pro-p group, the closed subgroup generated by this lift is ${\mathbb{Z}}_p$, and its map to $\Gamma$ is an isomorphism. This gives a semidirect product splitting.

The maximal abelian subextension $M/F$ is the fixed field of the closed commutator subgroup. Because $F_{\infty }/F$ is abelian, $F_{\infty }\subseteq M$. Restriction of the Galois correspondence to $X$ now gives the coinvariant formulaThe closed image above is important: the formula is a quotient of compact groups, not a quotient by an unclosed abstract subgroup. This is abelianization over a Zp-extension.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, where the bracket denotes the commutator subgroup. The group is soluble ifEquivalently it has a finite series with abelian factors. The trivial group is included.

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.