Involutions and Abelian Groups, II. In the thread Involutions and Abelian Groups, I supplied a solution to the following interesting problem (with the help of hints provided by the OP).

Let $ G $ be a finite group and $ I(G) $ the set of involutions of $ G $ (an involution of $ G $ is defined as an element $ x $ of $ G $ with order $ 2 $, i.e., $ x^{2} = e $). Prove that if $ \dfrac{|I(G)|}{|G|} \geq \dfrac{3}{4} $, then $ G $ must be an abelian group.

This problem naturally led me to consider the following intriguing question.

Can we find non-abelian finite groups $ G $ for which the ratio $ \dfrac{|I(G)|}{|G|} $ is less than but arbitrarily close to $ \dfrac{3}{4} $?

I am pretty sure that this problem has been studied extensively (and exclusively) by group theorists, but I have not been very successful in finding sources that discuss it. I do not even know if this question has an affirmative or negative answer. I therefore welcome all helpful suggestions and insights. Thank you!
 A: Here is a proof of my claim in the comments.
If $G$ is a nonabelian group of even composite order with $|I(G)|/|G|>1/2$, then $G$ is generalized dihedral by a theorem of Wall.  By a corollary of Frobenius-Schur (which appears as Corollary 4.6 in Isaacs Character Theory of Finite Groups), the number of involutions in $G$ is equal to $$-1+\sum_{\chi\in \text{Irr}(G)} \chi(1).$$
Our strategy will be to count the irreducible characters of each dimension using what we know about the structure of generalized dihedral groups, and thus determine the number of involutions in $G$.
Suppose that $G=A\rtimes \langle \tau \rangle$, where $A$ is abelian.  Then the quotient of $A$ by $N:=\{a^2|a\in A\}$ is an elementary abelian $2$-group.  Let $k=\text{rank}(A/N)$.  Each of the $2^k$ such characters from $A/N$ can send $\tau$ to either $\pm 1$, so Then $G$ has $2^{k+1}$ linear characters, as .  Next we can induce the characters of $A$ for which $A_2\not \leqslant \text{ker}(\chi)$ to obtain $(n-2^k)/2$ distinct characters of dimension $2$.
Letting $|A|=2^jm$ for some odd $m>2$, we have that $|G|=2^{j+1}m$, from which it follows that $$\frac{1+|I(G)|}{|G|}=\frac{1}{2^{j+1}m}\left(2^{k+1}+2\cdot \frac{(2^jm-2^k)}{2}\right)=\frac{1}{2}+\frac{1}{2^{1+j-k}m}$$ which is bounded above by $$\frac{1}{2}+\frac{1}{2m}$$ with equality when $j=k$, i.e. when the Sylow $2$-group of $A$ is elementary abelian.  This upper bound is a decreasing function of $m$, so the above calculation verifies that $|I(G)|/|G|$ approaches a maximum when we choose $m=3$, whereupon $$\frac{1}{2}+\frac{1}{2m}=\frac{2}{3}.$$
So Jacob's example $\mathbb{Z}_2^k\times S_3\cong D(\mathbb{Z}_3\oplus \mathbb{Z}_2^k)$ is in fact the maximal construction.
A: Consider the dihedral group $D_8$ of order $8$. Notice that it has $2$ elements of order $4$, $5$ elements of order $2$ and an identity element. Now consider the group $(\mathbb Z_2)^n \times D_8$. This is still non-abelian and we have that there are $2^n \cdot 5+2^n-1$ elements of order $2$ in particular we see that
$$\lim_{n\rightarrow \infty} \frac{6\cdot 2^n-1}{8\cdot 2^n}=3/4.$$
