Let $n_p$ be number of the elements of order $p$ in a group $G$.

My motivation is that if $n_2\ge\dfrac 34 |G|$ then $G$ is $2$-group. You can check it from this.

Is there such general bound for $n_p$ to conclude $G$ is a $p$-group?

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    $\begingroup$ Well, there is the somewhat trivial bound $n_p\ge |G|-1$ ... $\endgroup$ – Hagen von Eitzen Jun 10 '14 at 19:18
  • $\begingroup$ @HagenvonEitzen: as a constant ratio pleas :) $\endgroup$ – mesel Jun 10 '14 at 19:19
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    $\begingroup$ We would obviously want the best bound, so effectively we're searching for the supremum of $n_p/|G|$ over all non-$p$-groups $G$. Perhaps some numerical exploration could be done. $\endgroup$ – blue Jun 10 '14 at 19:21
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    $\begingroup$ As I pointed out in comment to the previous post about $n_2$, a Frobenius group with complement of order $p$ provides an example of a non $p$-group with $n_p = (p-1)|G|/p$, and you can increase the ratio slightly by taking a direct product with a large elementary abelian $p$-group. $\endgroup$ – Derek Holt Jun 10 '14 at 21:56
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    $\begingroup$ In case people are curious, here are some large values for $n_p/|G|$ I've found (where I take the supremum over $G \times C_p^n$, so these values are not attained in my examples, just approached): 2: 2/3, 3: 3/4, 5: 9/11, 7: 7/8, 11: 21/23, 13: 25/27, 17: 97/103, 19: 3611/3629. They are all from Frobenius groups as Derek suggested, and are best possible for $|G| < 500$ (larger $G$ give more opportunity for variety, but cannot take as much advantage of the $C_p^n$ trick). $\endgroup$ – Jack Schmidt Jun 11 '14 at 4:59

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