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.

Answer 1

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