What is the largest proportion of elements of a nonabelian group that have order at most $2$?
In other words, given $G$ a nonabelian group with order $n$, and $k$ elements of order at most $2$, how large can $k/n$ be?
A colleague and I have been thinking about the following for a couple of days. Unfortunately neither of us knows much about a group theory beyond a first course (and in particular, no representation theory to speak of), so it may be simple, but beyond our expertise. We have made some progress:
An abelian group of order $n$ can have $n$ elements of order at most $2$, at least when $n=2^m$ for some integer $m$: $(\mathbb{Z}/2\mathbb{Z})^m$ works.
Also, there is a priori no restriction on elements of prime order $p>3$ in nonabelian groups: a nonabelian group can have as many elements of order $p$ as we like: one of the two nonabelian groups of order $p^3$ has $p^3-1$ elements of order $p$, and this generalises to $p^m$ for $m>3$ by taking a direct product of this group with $(\mathbb{Z}/p\mathbb{Z})^{m-3}$.
Now, as for the case we're actually interested in: elements of order $2$ in nonabelian groups.
A lower bound for the maximum of the ratio $k/n$ is $3/4$: examples with this ratio are $ D_8 \times (\mathbb{Z}/2\mathbb{Z})^m $.
An upper bound for the maximum has been suggested by my colleague: suppose that $a,b$ are elements of order at most $2$. Then $$ ab = ba \iff aba = b \iff (ab)^2 = e, \tag{1} $$ i.e. $a$ commutes with $b$ if and only if $ab$ also has order at most $2$ (this immediately implies there must be an element of order larger than $2$, of course, or the group would be abelian). Now, the largest proportion of pairs that commute with one another is $5/8$ (i.e. there are at most $5n^2/8$ pairs that commute), by the $5/8$ theorem (John Baez blog post, MO post). If the set of elements of order at most $2$ is $H$ (not necessarily a subgroup), then $\lvert H \rvert = k$. Let $h \in H$, then $ hG = G $ by the group property. The most elements of $h' \in H$ for which the product $hh'$ can lie outside $H$ is $n-k$, so the number of $h'$s with $hh' \in H$ is at least $2k-n$, provided that $2k>n$ (the interesting case). There are $k$ $h$s, which is a total of at least $k(2k-n)$ pairs $hh'$s that lie in $H$. These pairs then all commute by $(1)$, so by the $5/8$ theorem, $$ k(2k-n) \leq \frac{5}{8} n^2 , $$ or $$ \left( \frac{k}{n}- \frac{1}{4} \right)^2 \leq \frac{3}{8} , $$ so $$ \frac{k}{n} \leq \frac{1}{4}(1+\sqrt{6}) < 0.863 . $$ One could do slightly better than this by treating the identity separately, worrying about elements with higher order that commute with their own powers, and so on, but none of this gives more than an $O(1/n)$ correction, as far as I can see.