Lemma 2

Let $H$ be a normal subgroup of $G$. There exists a subset $X$ of $S$ such that $BH = G_X$ and such that every element of $X$ commutes with every element of $S - X$.

Since $BH$ is a subgroup of $G$ containing $B$, there exists a unique subset $X$ of $S$ such that $BH = G_X$ (Th. 3).

Let $s_1 \in X$ and $s_2 \in S - X$; let $n_1$ and $n_2$ be representatives in $N$ of $s_1$ and $s_2$, respectively. Then $n_1 \in G_X = BH$ and there exists $b \in B$ such that $bn_1 \in H$. Since $H$ is normal in $G$, the element $h = n_2 b n_1 n_2^{-1}$ of $G$ belongs to $H$. This means that

If the length of $s_2 s_1 s_2$ is equal to 3, Cor. 1 of Th. 2 implies that

(6) If $H$ is a subgroup of a group $G$, the normaliser of $H$ in $G$ is the subgroup $\mathfrak{N}(H)$ consisting of the elements $g$ of $G$ such that $gHg^{-1} = H$. A subgroup $H'$ is said to normalise $H$ if $H' \subset \mathfrak{N}(H)$, in which case $HH' = H'H$ is a subgroup of $G$ in which $H$ is normal.

and hence that $h \in H \cap C(s_2 s_1 s_2)$. Since $H \cap C(s_2 s_1 s_2)$ is non-empty, $s_2 s_1 s_2 \in W_X$. As $(s_2, s_1, s_2)$ is a reduced decomposition, it follows that $s_2 \in X$, contrary to our assumption.

Thus $l_S(s_2 s_1 s_2) \leq 2$; if $l_S(s_2 s_1 s_2) = 1$, then $s_1 s_2 \in S$ and so $(s_1 s_2)^2 = 1$, or $s_1 s_2 = s_2 s_1$. If $l_S(s_2 s_1 s_2) = 2$, property (E) of §1, no. 5 implies that $s_2 s_1 = s_1 s_2$, since $s_1 \neq s_2$. Q.E.D.

The following property of a group U enters into Th. 5 below:

(R) For any normal subgroup V of U distinct from U, the commutator subgroup (cf. Algebra, Chap. I, §6, no. 8) of U/V is distinct from U/V.

Every soluble group satisfies (R); in particular, every abelian group satisfies (R). It can be shown that the symmetric group $\mathfrak{S}_n$ satisfies (R) (cf. Exerc. 29).