5
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ $D_8$ 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
4
$\begingroup$

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 $2$

$\endgroup$
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.