Suppose $n_p(G)$ is the number of elements of order $p$ in a group $G$.
Does there for every prime $p$ exist a $\epsilon_p > 0$, such that for any group $G$ $n_p(G) > (1 - \epsilon_p(G))|G|$ implies, that $G$ is a $p$-group?
This question was inspired by a well known fact, that for any group $G$ $n_2(G) > \frac{3}{4}|G|$ implies $G$ being a $2$-group (or, alternatively, that $\epsilon_2 = \frac{1}{4}$).
Yes, this is the main theorem of
Laffey, Thomas J., The number of solutions of (x^p=1) in a finite group, Math. Proc. Camb. Philos. Soc. 80, 229-231 (1976). ZBL0343.20006.
Namely, if a finite group $G$ has more than $\tfrac{p}{p+1} |G|$ elements of order $p$, then $G$ is a $p$-group. This bound is achieved by $C_p \ltimes C_2^k$ where $p = 2^k-1$ is a Mersenne prime.
There is a related very frustrating problem: Is there an $\delta_p$ such that, if $G$ has more than $1-\delta_p$ elements of order $p$, then $G$ is $p$-torsion? We can take $\delta_3 = \tfrac{2}{9}$, see
Laffey, Thomas, The number of solutions of (x^3=1) in a 3-group, Math. Z. 149, 43-45 (1976). ZBL0314.20020.
I thought for a bit that the right bound might be $\tfrac{p-1}{p^2}$ in general, which occurs for $C_p \ltimes \mathbb{Z}[\zeta_p]/(1-\zeta_p)^N$ where $\zeta_p$ is a $p$-th root of unity and $p \leq N \leq 2p-2$. But that's wrong! Wall constructed a $5$-group of exponent $25$ where all the elements of order $25$ were in an index $25$ subgroup, demonstrating that $\delta_5$, if it exists, is $< \tfrac{1}{25}$.
Wall, G. E., On Hughes’ $H_{p}$ problem, Proc. Int. Conf. Theory Groups, Canberra 1965, 357-362 (1967). ZBL0189.31701
But, as far as I know, no one knows whether any such $\delta_5$ exists at all!
See Havas, George; Vaughan-Lee, Michael, On counterexamples to the Hughes conjecture., J. Algebra 322, No. 3, 791-801 (2009). ZBL1187.20010
for groups achieving $\tfrac{1}{p^2}$ for various $p>5$. I will also note that I talked to Harry Altman about a lot of this material.
