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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

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    $\begingroup$ $D_8$ has 5 involutions. $\endgroup$ – Myself Feb 14 '15 at 12:43
  • $\begingroup$ Hint half the elements in a dihedral group are reflections. What if there is an even number of rotations? $\endgroup$ – Tobias Kildetoft Feb 14 '15 at 12:45
  • $\begingroup$ symmetric group $S_3$ has exactly half of numbers with this property. $\endgroup$ – Bumblebee Apr 28 '15 at 9:41
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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 $2$

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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

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