Subgroup Generated by Commuting Elements is Abelian

Theorem

Let $\struct {G, \circ}$ be a group.

Let $S \subseteq G$ such that:

$\forall x, y \in S: x \circ y = y \circ x$

Then the subgroup generated by $S$ is abelian.

Proof

Let $H = \gen S$ denote the subgroup generated by $S$.

Let $a, b \in H$.

Then:

$a = s_1$

$b = s_2$

for some words $s_1, s_2$ of the set of words $\map W S$ of $S$.

Then:

\(\ds a \circ b\)

\(=\)

\(\ds s_1 \circ s_2\)

\(\ds \)

\(=\)

\(\ds s_2 \circ s_1\)

as all elements of $S$ commute with each other

\(\ds \)

\(=\)

\(\ds b \circ a\)

Hence the result, by definition of abelian group.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $5$: Subgroups: Exercise $9$