Proposition 27

Let $\mathbf{N}$ and $\mathbf{L}$ be two groups, $e'$ and $e''$ their respective identity elements. Suppose that we are given a homomorphism $y \to \sigma_y$ of $\mathbf{L}$ into the group $\Gamma$ of automorphisms of $\mathbf{N}$. Then:

1) On the product set $S = \mathbf{N} \times \mathbf{L}$, the internal law of composition

defines a group structure, for which $j_1 : x \to (x, e'')$ is an isomorphism of $\mathbf{N}$ onto a normal subgroup of $S$, $j_2 : y \to (e', y)$ is an isomorphism of $\mathbf{L}$ onto a subgroup of $S$, and $\mathrm{pr}_2 : S \to \mathbf{L}$ is a surjective homomorphism whose kernel is $j_1(\mathbf{N})$ and which is such that $\mathrm{pr}_2 \circ j_2$ is the identity automorphism of $\mathbf{L}$.

2) Let $f : \mathbf{N} \to G$, $g : \mathbf{L} \to G$ be two homomorphisms into a group $G$, such that

for all $x \in \mathbf{N}$ and all $y \in \mathbf{L}$. Then there is a unique homomorphism $h : S \to G$ such that $f = h \circ j_1$, and $g = h \circ j_2$.

If $x, x', x''$ are elements of $\mathbf{N}$ and $y, y', y''$ are elements of $\mathbf{L}$, then

and and therefore the associativity of the law (3) follows from the facts that $y \to \sigma_y$ is a homomorphism of $\mathbf{L}$ into $\Gamma$ and that $\sigma_y$ is an automorphism of $\mathbf{N}$. Clearly $(e', e'')$ is the identity element of (3), and finally so that $(x, y)$ has an inverse in $S$. The other assertions of 1) are clear. On the other hand, since $(x, y) = (x, e'') (e', y)$, a homomorphism $h$ which satisfies the conditions of 2) must necessarily satisfy hence is unique if it exists; moreover it is immediate from (4) that

which shows that $(x, y) \to f(x)g(y)$ is indeed a homomorphism of S into G satisfying the conditions of 2).