A lot of even elements

Back when I was an undergraduate we were asked to name a finite noncommutative group such that more then half the elements have order two. Furthermore, we know that every group such that every element different from the identity has order two must be abelian. So I ask myself, "how much of an noncommutative group can be of order two". In a more rigorous way:

What do I get, when I take the supremum over all finite noncommutative groups $G$ of the following expression $$ \frac{\vert\{ g\in G \ : \ g \text{ has order } 2\}\vert}{\vert G \vert}.$$

I was able to show that the supremum should be greater equal $2/3$ by considering the groups $G_n =(\mathbb{Z}/2\mathbb{Z})^n \oplus S_3$ which has $4\cdot 2^n -1$ elements of order two and $\vert G_n \vert = 6\cdot 2^n$.

This preprint has a self-contained proof that the answer is $\frac34$:

Allan L. Edmonds and Zachary B. Norwood (2009) "Finite groups with many involutions" https://arxiv.org/abs/0911.1154

It is shown that a finite group in which more than 3/4 of the elements are involutions must be an elementary abelian 2-group. A group in which exactly 3/4 of the elements are involutions is characterized as the direct product of the dihedral group of order 8 with an elementary abelian 2-group.

(I'll leave this answer as community wiki in case someone else wants to read the paper and summarize the argument.)