In this, involutions in a finite group are either conjugate or have an involution centralizing both of them. I wonder if there are similar results for an infinite group. I think and look for it but I can't finish it. When I read this proof, I wonder if a product of two involutions can be written as a commutator. Thanks for all your support.
Involution elements in infinite groups
abstract-algebragroup-theoryinfinite-groupsinvolutions
Related Solutions
Simple examples are given by free products, assuming you know the normal form for a free product; otherwise, I'm not really saying much.
If you want a more concrete example, take the $2\times 2$ matrices with coefficients in $\mathbb{Q}$, and $$a = \left(\begin{array}{cc}0&1\\1&0\end{array}\right),\qquad b=\left(\begin{array}{cc}0 & 2\\\frac{1}{2}& 0\end{array}\right).$$ Then $a^2=b^2=1$, but $$ab = \left(\begin{array}{cc}\frac{1}{2} & 0 \\0 & 2\end{array}\right)$$ has infinite order.
If $a$ and $b$ commute, with $\mathrm{ord}(a)=m$ and $\mathrm{ord}(b)=n$, then you can do better than Wikipedia. We have that: $$\frac{\mathrm{lcm}(m,n)}{\mathrm{gcd}(m,n)}\quad\text{divides}\quad \frac{\mathrm{lcm}(m,n)}{|\langle x\rangle\cap\langle y\rangle|}\quad\text{divides}\quad \mathrm{ord}(ab)\quad\text{divides}\quad \mathrm{lcm}(m,n).$$ For example, if for every prime $p$ that divides $\mathrm{lcm}(m,n)$, the highest power of $p$ that divides $m$ is different from the highest power of $p$ that divides $n$, then $\mathrm{ord}(ab)=\mathrm{lcm}(m,n)$.
If you are willing to impose global conditions (conditions on $G$), then for example Easterfield proved in 1940 that if $G$ is a $p$-group of class $c$, $a$ has order $p^{\alpha}$, and $b$ has order $p^{\beta}$, then the order of $ab$ is at most $p^m$, where $$ m = \max\left\{\alpha,\beta+\left\lfloor\frac{c-1}{p-1}\right\rfloor\right\}.$$ From this you can get similar results for a finite nilpotent group (which is necessarily a product of $p$-groups), and hence for the product of two elements of finite order in any nilpotent group (or in fact, in any locally nilpotent group, or even more strongly, in any group in which the 2-generated subgroups are nilpotent).
No, your proof does not work. In general the number of involutions in a group is not equal to $\frac{|G|}{2}+1$ (btw this is not an odd number in general). See LINK for the case of the symmetric group (you have to subtract $1$, since there the identity $e$ is considered an involution).
Hint. By Lagrange's theorem, a group containing an element of order 2, i.e. a strict involution, has even order. Now $G$ can be split into three disjoint sets: $$G=\{e\}\cup\{\text{elements of $G$ of order$=2$}\}\cup \{\text{elements of $G$ of order$>2$}\}.$$ Notice that if $g$ has order $>2$ then $g\not=g^{-1}$ which implies that the set $\{\text{elements of $G$ of order$>2$}\}$ can be furtherly split into two sets of the same cardinality.
Can you take it from here?
Best Answer
Let $x,y$ in a group satisfy $x^2=y^2=1$. Then (as David Craven wrote in an erased comment) one has one of the following:
Indeed, if $xy$ has finite odd order, then (1) holds. If $xy$ has finite even order, then (2) holds. And otherwise (3) holds.
In addition, no condition can be removed, as generators in various dihedral groups show. Also, in suitable groups and pairs, any two of these conditions can be satisfied without the third, or all three simultaneously.