TITLE: Non-abelian finite group in which more than half of the elements have order $2$
QUESTION [5 upvotes]: Is there an non-abelian finite group, in which more than half of the elements have order $2$

I only know that if there is one, then all elements (except identity) cannot have order $2$, otherwise it would be abelian, so there is at least one element of order $>2$, say $x$. If I conjugate $x$ with another element, then conjugation preserves order but any $2$ conjugation might not give distinct elements, for the conjugacy class of an arbitrary element would be always the whole group.
Or is there such a group

REPLY [5 votes]: In fact, the threshold to force the group to be elementary abelian is 3/4, see 
http://arxiv.org/abs/0911.1154
and also
https://mathoverflow.net/questions/40028/half-or-more-elements-order-two-implies-generalized-dihedral