Theorem 4

Let G be a group with operators and N a normal stable subgroup of G.

(a) The mapping $G' \mapsto G'/N$ is a bijection of the set of stable subgroups of G containing N onto the set of stable subgroups of G/N.

(b) Let $G'$ be a stable subgroup of G containing N. For $G'/N$ to be normal in $G/N$, it is necessary and sufficient that $G'$ be normal in G and the groups $G/G'$ and $(G/N)/(G'/N)$ are then isomorphic.

(c) Let $G'$ be a stable subgroup of G. Then $G'N$ is a stable subgroup of G and N is normal in $G'N$. Further $G' \cap N$ is normal in $G'$ and the groups $G'/(G' \cap N)$ and $G'N/N$ are isomorphic.

Let $f$ denote the canonical homomorphism of G onto $G/N$. For all $x \in G$, $f(x) \in xN$; therefore, $f(G') = G'/N$ for every subgroup $G'$ of G containing N. As $f$ is surjective, assertion (a) follows from Corollary 1 to Proposition 7; similarly for the equivalence "G' normal" $\Leftrightarrow$ "G'/N normal". Suppose that $G'$ is a normal stable subgroup of G containing N. By no. 4, Proposition 5 applied to $\mathrm{Id}_G$, there exists a homomorphism $u$ of $G/N$ into $G/G'$ defined by $u(xN) = xG'$ for all $x \in G$. It is immediate that $u$ is surjective with kernel $G'/N$ whence the desired isomorphism of $(G/N)/(G'/N)$ onto $G/G'$. Finally, (c) follows immediately from Proposition 8.