# Can we find a bound so that we can conclude $G$ is a $p$-group?

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?

• Well, there is the somewhat trivial bound $n_p\ge |G|-1$ ... – Hagen von Eitzen Jun 10 '14 at 19:18
• @HagenvonEitzen: as a constant ratio pleas :) – mesel Jun 10 '14 at 19:19
• 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. – blue Jun 10 '14 at 19:21
• 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. – Derek Holt Jun 10 '14 at 21:56
• 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). – Jack Schmidt Jun 11 '14 at 4:59