Proposition 5

Let $A, B, C$ be three subgroups of $G$.

(i) The subgroup $A$ normalizes the subgroup $(A, B)$.

(ii) If the subgroup $(B, C)$ normalizes $A$, the subgroup $(A, (B, C))$ is generated by the elements $(a, (b, c))$ with $a \in A, b \in B$ and $c \in C$.

(iii) If $A, B$ and $C$ are normal, then

By (4 bis), for $a, a' \in A$ and $b \in B$,

whence (i). Suppose now that $(B, C)$ normalizes $A$. For $a \in A, b \in B, c \in C$ and $x \in G$, (4) implies

and $((b, c), a) \in A$ since $(B, C)$ normalizes $A$, whence by induction on $p$ the fact that $\left( a, \prod_{i=1}^p (b_i, c_i) \right)$, for $b_i \in B, c_i \in C$, belongs to the subgroup generated by the elements of the form $(a, (b, c))$. If finally $A, B$ and $C$ are normal, so are the subgroups $(A, (B, C))$, $(C, (B, A))$ and $(B, (C, A))$. It therefore suffices by (ii) to show that

for all $a \in A, b \in B$ and $c \in C$. Now by (5), writing $a^{b^{-1}} = u$

whence (iii).