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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.

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2 Answers

up vote 15 down vote accepted

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.

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Another 'yes' answer can be found through the Tarski Monster groups.

A group $G$ is a Tarski Monster $p$-group for some prime $p$ if $G$ is infinite, finitely generated, and every proper non-trivial subgroup of $G$ is cyclic of prime order. Such a group is necessarily simple (see below). Ol'shanskii proved that these existed for all "$p$ large enough" ($p>10^{75}$ or something similar) in a series of papers in the late 70s-early 80s. He published a book which contains the proof in '89 (Russian)/'91 (English), called The geometry of defining relations in groups.

Mark Sapir and others have done work recently on Monster groups in general. Ol'shanskii's construction is infinitely presented (but two-generated), and I believe it is still an open problem as to whether there exist finitely presented Tarski Monsters.


Below: Let $T$ be a Tarski Monster group. Then we prove that $T$ is simple.

Assume $T$ is not simple. Thus, there exists some $K$ such that $\langle 1\rangle\neq K\lhd T$. By the definition of $T$, $K=\langle x\rangle\cong C_p$, and as $K$ is normal we have that for arbitrary $g\in T$, $g$ acts on $K$ by conjugation. Thus, $\langle g\rangle\rightarrow \operatorname{Aut}(C_p)$. However, $\langle g\rangle\cong C_p$ while $|\operatorname{Aut}(C_p)|=\varphi(p)=p-1$ (in fact, it is well known that $\operatorname{Aut}(C_p)=C_{p-1}$) so the action of $g$ on $K$ is trivial. That is, $gxg^{-1}=x$ for all $g\in G$.

Now, let $g\in G\setminus K$ be arbitrary. Then either $\langle x, g\rangle=T$ or $\langle x, g\rangle\cong C_p$. However, $x$ and $g$ commute while $T$ is non-abelian so $\langle x, g\rangle\neq T$. On the other hand, $C_p=\langle x\rangle\lneq \langle x, g\rangle$ so $\langle x, g\rangle\not\cong C_p$. This is a contradiction. Thus, $T$ is simple.

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I think you can also see it this way. If $K$ is normal, then pick a proper subgroup $L$ that is not $K$. Then $KL$ has order $p^2$. –  Mikko Korhonen May 2 '12 at 14:51
    
@m.k.: That's much neater. Thanks. –  user1729 May 2 '12 at 15:01
    
@user1729: I should analyze your neat answer carefully. +1 –  B. S. Dec 10 '12 at 12:07
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