# No Group has Two Order 2 Elements

## Theorem

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

## Proof

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Suppose:

- $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$

That is. they are self-inverse:

- $s^2 = e = t^2$

Hence $s$ nor $t$ is the identity

Hence, as $s \ne t$, 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 $\paren {s \circ t}^2 = e$ from Self-Inverse Elements Commute iff Product is Self-Inverse.

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 $G$ distinct from both $s$ and $t$.

But from Order of Conjugate Element equals Order of Element:

- $\order {s \circ t \circ s^{-1} } = \order t$

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

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $12 \ \text {(ii)}$