Proposition 8

Let $G$ be a group with operators and $A$ and $B$ two stable subgroups of $G$. Suppose that the relations $a\in A$ and $b\in B$ imply $aba^{-1}\in B^{*}$ (in other words, $A$ normalizes $B$)$^{*}$. Then $AB=BA$ is a stable subgroup of $G$, $A\cap B$ is a normal stable subgroup of $A$ and $B$ is a normal stable subgroup of $AB$. The canonical injection of $A$ into $AB$ defines on passing to the quotient an isomorphism of $A/(A\cap B)$ onto $AB/B$.

The formulae

for $a,a'\in A$, $b,b'\in B$ and every operator $\alpha$ on $G$, show that $AB$ is a stable subgroup of $G$. Let $a\in A$ and $x\in A\cap B$; then $axa^{-1}\in B$ by the hypotheses made on $A$ and $B$ and clearly $axa^{-1}$ belongs to $A$, hence $A\cap B$ is normal in $A$. Let $a\in A$ and $b,b'$ be in $B$; the formula $(ab)b'(ab)^{-1}=a(bb'b^{-1})a^{-1}$ shows that $B$ is normal in $AB$. Let $\phi$ be the restriction to $A$ of the canonical homomorphism of AB onto AB/B; then $\phi(a) = aB$ and hence the kernel of $\phi$ is equal to $A \cap B$. Clearly $\phi$ is surjective and hence defines an isomorphism of $A/(A \cap B)$ onto $AB/B$.