All Elements Self-Inverse then Abelian

Theorem

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

Suppose that every element of $G$ is self-inverse.


Then $G$ is abelian.


Proof

Every element of $G$ is self-inverse, that is:

$\forall x \in G: x \circ x = e$

In particular, for all $x, y \in G$:

$\paren {x \circ y} \circ \paren {x \circ y} = e$

that is, $x \circ y$ is also self-inverse.


From Self-Inverse Elements Commute iff Product is Self-Inverse, it follows that:

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

that is: $x$ and $y$ commute for all $x, y \in G$.


Hence $G$ is an abelian group.

$\blacksquare$


Sources

  • 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$: Exercise $2.2$
  • 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{T}$
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 29 \delta$
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