Yes, this is the main theorem of 

<cite authors="Laffey, Thomas J.">_Laffey, Thomas J._, [**The number of solutions of \(x^p=1\) in a finite group**](http://dx.doi.org/10.1017/S0305004100052865), Math. Proc. Camb. Philos. Soc. 80, 229-231 (1976). [ZBL0343.20006](https://zbmath.org/?q=an:0343.20006).</cite>

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.

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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 

<cite authors="Laffey, Thomas">_Laffey, Thomas_, [**The number of solutions of \(x^3=1\) in a 3-group**](http://dx.doi.org/10.1007/BF01301629), Math. Z. 149, 43-45 (1976). [ZBL0314.20020](https://zbmath.org/?q=an:0314.20020).</cite>

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 $N \geq p$. But that's wrong! For $p=5$, the construction of

<cite authors="Wall, G. E.">_Wall, G. E._, On Hughes’ \(H_{p}\) problem, Proc. Int. Conf. Theory Groups, Canberra 1965, 357-362 (1967). [ZBL0189.31701](https://zbmath.org/?q=an:0189.31701)</cite>

shows that we can't beat $\tfrac{1}{25}$. 

But, as far as I know, no one knows whether any such $\delta_5$ exists at all! 

See 
<cite authors="Havas, George; Vaughan-Lee, Michael">_Havas, George; Vaughan-Lee, Michael_, [**On counterexamples to the Hughes conjecture.**](http://dx.doi.org/10.1016/j.jalgebra.2009.04.011), J. Algebra 322, No. 3, 791-801 (2009). [ZBL1187.20010](https://zbmath.org/?q=an:1187.20010)</cite>

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.