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.