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 Commute iff Product 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 $G$ 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 $2$.