According to the Feit-Thompson theorem, every group of odd order is solvable and thus every finite nonabelian simple group has even order. Thus every finite nonabelian simple group has an involution (element of order $2$).
My question is the following: is there an infinite simple group that has no element of order $2$?
This was just something I thought about, so I have no idea how hard this question is. So a complete answer might go over my head, but anyone answering this shouldn't worry since others might find the answer valuable. I'm also interested in just knowing whether this is true or not.
Best Answer
Yes, because there exist torsion-free infinite simple groups (i.e. simple groups having no elements of finite order).
Here you can read a bit about one type of example of such groups.