Lemma 7

Let B be a group, A a normal subgroup of B, C the group $B/A$ and $i : A \to B$ and $p : B \to C$ the canonical morphisms. Let $A'$ be a group and $f$ a homomorphism of A into $A'$. Let $\omega$ be a morphism of B into the automorphism group of $A'$. Suppose that, for $a \in A'$, $a' \in A'$, $b \in B$,

Let $B''$ be the semi-direct product of B by $A'$ relative to $\omega$ and q the canonical morphism of $B''$ onto B.

(i) The mapping $a \mapsto (f(a^{-1}), i(a))$ of A into $B''$ is a morphism of A onto a normal subgroup D of $B''$. Let $B' = B''/D$ and $i' : A' \to B'$, $g : B \to B'$ be the morphisms of $A'$ and B into $B'$ derived when taking quotients from the canonical injections of $A'$ and B into $B''$.

(ii) The morphism $p \circ q$ of $B''$ into C defines when passing to the quotient a morphism of $B'$ into C.

(iii) $i'$ is injective, $p'$ is surjective, $\mathrm{Ker}(p') = \mathrm{Im}(i')$ and the following diagram is commutative

(iv) If $b \in B$ and $a' \in A'$, then

(i) For $a_1, a_2$ in $A$, we have, in $B''$,

and hence $a \mapsto (f(a^{-1}), i(a))$ is a homomorphism $h$ of $A$ into $B''$. Let $a \in A$, $a' \in A'$, $b \in B$; then, in $B''$, and hence $h(A) = D$ is normal in $B''$.

(ii) For $a \in A$,

and hence $p \circ q$ is trivial on $D$.

(iii) Let $a' \in A'$ be such that $i'(a') = e$; then $a' \in D$ and hence there exists $a \in A$ such that $a' = f(a^{-1})a$; this implies $a = e$, whence $a' = e$; thus $i'$ is injective. As $p$ and $q$ are surjective, $p'$ is surjective.

Let $r$ denote the canonical morphism of $B''$ onto $B'$. Let $a' \in A'$, $b \in B$ and $b' = r(a'b)$. If $b' \in \mathrm{Im}(i')$, there exists $a'_1 \in A'$ such that $b' = r(a'_1)$; then there exists $a \in A$ such that $a'b = a'_1f(a^{-1})a$, whence $b = a \in A$ and

thus, $\mathrm{Im}(i') \subset \mathrm{Ker}(p')$. We preserve the notation $a', b, b'$ but assume that $b' \in \mathrm{Ker}(p')$; then $e = p'(b') = p(q(a'b)) = p(b)$ and hence $b \in A$, whence thus $\mathrm{Ker}(p') \subset \mathrm{Im}(i')$.

If $a \in A$, then

If $b \in B$, then

and hence diagram (4) is commutative.

(iv) Let $b \in B, a' \in A'$. Then