Is it necessarly abelian $2$ group?

Question

Is the following correct ?

Let $n_2$ be a number of elements of $G$ of order $2$ if $n_2> \dfrac {|G|}2$ then $G$ is elemantary abelian $2$ group.

Edit: we see that the conclusion "if $n_p> \dfrac {|G|}p$ then $G$ is a $p$ group is wrong" thanks to JyrkiLahtonen.

Even though it's been answered, you should leave the second question in the OP. — , Jun 10 '14 at 18:38
Oh well. Then I need to retype the comment telling that $D_5$ is a counterexample :-) — Jyrki Lahtonen, Jun 10 '14 at 18:42

score 7 · Accepted Answer · edited Apr 13 '17 at 12:21

7

The direct product of an elementary abelian group of order $2^n$ with a dihedral group of order $8$ has $3 \times 2^{n+1} - 1$ involution, so the proportion is $3/4 - 2^{-(n+3)}$.

It is a known result that if at least $3|G|/4$ elements are involutions, then $G$ is elementary abelian. See Involutions and Abelian Groups

edited Apr 13 '17 at 12:21

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answered Jun 10 '14 at 18:44

Derek Holt

96,726

Thanks, Do you know any other bound for $n_p$ to get $G$ as an $p$ group ? – mesel Jun 10 '14 at 18:51
1

I don't know of any bounds for $n_p$. But, for any prime, a Frobenius group with complement $C_p$ will have $n_p/|G| = (p-1)/p$, and you can increase that slightly by taking a direct product with a big elementary abelian $p$-group. So, for $p=3$, by taking $A_4 \times E$, I can get arbitrarily close to $3/4$ for $n_p/|G|$. I don't know if anyone can beat that or prove it's the best possible. – Derek Holt Jun 10 '14 at 19:39

Is it necessarly abelian $2$ group?

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