No Group has Two Order 2 Elements

Theorem

A group can not contain exactly two elements of order $2$.

Proof

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Suppose $s, t \in \left({G, \circ}\right): s \ne t, \left|{s}\right| = \left|{t}\right| = 2$.

That is, $s^2 = e = t^2$, i.e. they are Self-Inverse.

As $s \ne t$, and neither $s$ nor $t$ is the identity (as the identity is of order $1$), then $s \circ t \in G$ is distinct from both $s$ and $t$.

Also $s \circ t \ne e$ because $s \ne t^{-1}$.

• Suppose $s$ and $t$ commute. Then $\left({s \circ t}\right)^2 = e$ from Self-Inverse Elements that Commute.

Thus there is a third element (at least) in $G$ which is of order $2$.

• Now suppose $s$ and $t$ do not commute.

Then from Commutation Property in Group, $s \circ t \circ s^{-1}$ is another element of $G$ different from both $s$ and $t$.

But from Order of Conjugate, $\left|{s \circ t \circ s^{-1}}\right| = \left|{t}\right|$, and thus $s \circ t \circ s^{-1}$ is another element of order $2$.

Thus there is a third element (at least) in $G$ which is of order $2$.

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $6.12 \ \text {(ii)}$