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Consider the following problem, appropriate for a first course in Group Theory:

Problem: Prove that there cannot be a group with exactly two elements of order $2$.

General Proof: Suppose for the sake of contradiction that there are exactly two elements of order $2$, and denote them by $a$ and $b$. Note $ab \neq a, b, e$ so that $(ab)^2 \neq e$ or else we'd have a third element of order $2$. From here, one can show that $aba \neq a, b, e$, and then observe that $(aba)^2 = e$, giving a third element of order $2$. Contradiction.

Finite Group Proof: Given a finite group $G$, we can prove the statement as follows:

Suppose $a, b \in G$ are the only elements with order $2$. Observe $\{e, a\} \subset G$ is a subgroup with $2$ elements, so (by Lagrange's Theorem) $G$ has an even number of elements.

Create the Cayley table for $G$ and consider the NW/SE diagonal, on which $e$ will appear exactly three times: $e*e, a*a,$ and $b*b$. For any other elements whose product is $e$, they will appear in the table in pairs: if $x*y = e$ for $x \neq y$ then $y*x = e$ as well.

Thus, the total number of $e$'s in the table is odd: the three in the diagonal plus the even number that come in pairs. But the number of $e$'s in the table should be even: one for each row, and we proved $G$ has an even number of elements. Contradiction.

Question: Can the latter approach be adapted to cover the case for infinite groups?

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up vote 2 down vote accepted

One can show that the subgroup generated by $a$ and $b$ is finite, so that the problem is reduced to the finite case. Then, the same argument applies.

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Conjugating $b$ by $a$ basically brings things back to the General Proof, which I had (perhaps unrealistically) hoped to avoid. Nonetheless, I will accept this answer unless someone figures out how to somehow skirt conjugation... – Benjamin Dickman Oct 27 '12 at 0:12

You can show even more, namely:

If a group $G$ contains finitely many involutions, then it contains an odd number of them.

Here, an involution is an element of order 2.

The finite case is very well-known, and follows by trying to pair elements up with their inverses (and using the fact $|G|$ is even).

The infinite case then follows from the finite case by using Dietzmann's lemma, which says that if $X\subset G$ is a finite set of finite-order elements, and $X$ is closed under conjugation in $G$, then $\langle X\rangle$ is a finite subgroup of $G$.

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