# Non-abelian finite group in which more than half of the elements have order $2$

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

• $D_8$ has 5 involutions. – Myself Feb 14 '15 at 12:43
• Hint half the elements in a dihedral group are reflections. What if there is an even number of rotations? – Tobias Kildetoft Feb 14 '15 at 12:45
• symmetric group $S_3$ has exactly half of numbers with this property. – Bumblebee Apr 28 '15 at 9:41

Consider the group of symmetries of a square. $\mathbb{D}_4$. Then, flip along horizontal, vertical and the two diagonals are of order $2$. Also, rotation by 180° is of order $2$