# Is the order of the product of two commutative finite-order elements necessarily finite?

In some group G, can we exhibit an example of two elements $x,y$ that

• commute with each other
• have finite order

but whose product $xy$ (or $yx$, since they commute) have infinite order?

I can give an example of when the elements don't commute, say of the permutation group on countable digits, and by defining $f(x) = 1 -x$, $g(x) = 2-x$. Then $f \circ g$ and $g \circ f$ are both infinite order.

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## 1 Answer

If $x$ has order $n$, $y$ has order $m$ and $xy=yx$ then $o(xy) \mid nm$. This follows from observing that $$(xy)^{nm}=x^{nm}y^{nm}=e^me^n=e.$$

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2
Argh, my bad. I fretted unnecessarily over the possibility that the powers of (say) $a$ may not commute with $b$! Thanks for the quick answer. I'll accept your answer in a bit as the system is not letting me do that now. – math1793 Oct 11 '12 at 4:38
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