Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?
Wikipedia says "not much":
There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.
On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:
- What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?
Wikipedia then states:
If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$
which is easy to prove, but not very effective. So:
- What are some similar results about the order of a product, perhaps with some additional hypotheses?
Best Answer
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).