Other method(s) to prove "a group cannot have exactly two elements of order $2$" If $a,b$ are elements of a group having order $2$ then, if $a,b$ commute, $ab(\ne a , \ne b)$ has order $2$, and if $a,b$ do not commute, then $aba^{-1}(\ne a , \ne b, \ne e)$ has order $2$. Using this we can deduce "a group cannot have exactly two elements of order $2$".

Is there any other method to prove this statement?

 A: A somewhat complicated way consists in noting that in any group, the subgroup generated by two distinct involutions is dihedral, of order at least $4$ (possibly infinite). And in a dihedral group there are plenty of involutions.
A: If the group is finite, more is true. From the McKay's proof of Cauchy's theorem for $p = 2$, the number of elements of order two is odd. Hence, it is not two.
A: You can make McKay's proof work in the general (not necessarily finite) case by referencing Dietzmann's Lemma.  Namely, if any group has a finite number of order $p$ elements, the number of such elements is congruent to $-1\pmod{p}$.
A: Suppose that the group $G$ contains only two distinct elements of order $2$ say $a$ and $b.$ It is well-known that $D = \langle a,b \rangle$ is a (possibly infinite) dihedral group ( if this is unfamiliar, note that $b(ab)b^{-1} = ba = (ab)^{-1} ).$ Hence $C \lhd D$, where $C = \langle ab \rangle.$ Now $bcb^{-1} = c$ for all $ c \in C,$ so that $bcbc = 1$ for all $c \in C.$ Note that $bc \in D \backslash C$ for all $c \in C,$ so that $bc \neq 1$ for any $c \in C.$ Hence we must have $bc \in \{a,b \}$ for all $c \in C.$ If $bc =1,$ then $c = b,$ while if $bc = a$, then $c = ba \neq 1.$ Hence we must have $|C| = 2$ and $|D| = 4.$ But then $D$ ia Abelian,and $ab = ba$ is a third element of order $2$ in $G,$ a contradiction.
(But the proof given in the text of the question is better and not essentially different!).
A: To fill in a missing detail from other answers, the subgroup $D = \langle a,b\rangle$ of $G$ generated by the two involutions $a,b$ is dihedral, for the following reason. The free-est possible group generated by two involutions is the infinite dihedral group, which is a free product $$D_\infty = (\mathbb{Z}/2\mathbb{Z}) * (\mathbb{Z}/2\mathbb{Z}) = \langle a,b \mid a^2 = b^2 = \text{Id}\rangle
$$
There is a semidirect product structure $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$ where the kernel $\mathbb{Z}$ is generated by $ab$ and where the kernel $\mathbb{Z}/2$ acts on $\mathbb{Z}$ by transposing the generator $ab$ with its inverse $ba$. From this one easily deduces that every order two element of $D_\infty$ is conjugate to either $a$ or $b$. Also, the only normal subgroups of $D_\infty$ not containing $a$ or $b$ are subgroups of the kernel, which means the trivial group or the infinite cyclic group generated by some power $(ab)^k$. Thus, the only quotients of $D_\infty$ in which $a,b$ are nontrivial and not equal are $D_\infty$ itself, and the finite dihedral groups $\langle a,b \mid a^2 = b^2 = (ab)^k \rangle$ with $k \ge 2$. But each such group contains a third involution, such as $bab$.
