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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?

the_fox
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mesel
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    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
  • OK, then the trivial bound is $n_p\ge 1\cdot |G|$ which is vacuously correct :) – Hagen von Eitzen Jun 10 '14 at 19:20
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    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. – anon Jun 10 '14 at 19:21
  • I have only one trivial observation, if $S_p$ be set of elements of order $p$ and if $S_p>\dfrac {|G|}{2}$ then all other elements can be written product of two elemets of order $p$. – mesel Jun 10 '14 at 19:29
  • And another observation $$n_p\equiv 0 \ mod(p-1)$$ $$n_p\equiv -1 \ mod(p)$$ – mesel Jun 10 '14 at 19:31
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    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
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    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 04:59
  • @Jack Schmidt, Are you sure about the value for $p=2$? A group with more than half of its elements of order $2$ must be a $2$-group, so the best value for $n_2$ is $1/2$. (See for example the references in http://mathoverflow.net/questions/40028/half-or-more-elements-order-two-implies-generalized-dihedral) – verret Feb 02 '16 at 07:32

1 Answers1

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Yes such general bound actually does exist. It was proved by Thomas J. Laffey in "The number of solutions of $x^p=1$ in finite groups". That theorem states:

If $n_p > \frac{p}{p + 1} |G|$ then $G$ is a $p$-group

Note, that for $p = 2$ this theorem yields a result, stronger, than yours (namely $\frac{2}{3}$ instead of $\frac{3}{4}$)

Chain Markov
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