From Joseph A. Galligan's Contemporary Abstract Algebra (2nd ed.), we have the following corollary:
Corollary 3 (Relationship between $|ab|$ and $|a||b|$). If $a$ and $b$ belong to a finite group and $ab = ba$, then $|ab|$ divides $|a||b|$.
Proof. Let $|a| = m$ and $|b| = n$. Then $(ab)^{mn} = (a^m)^n(b^n)^m = e^n e^m = e$. So, by Corollary 2 of Theorem 4.1 we have that $|ab|$ divides $mn$.
For reference, Corollary 2 says that $a^k = e$ implies $|a|$ divides $k$.
This proof makes sense in all areas except for one—each step in the proof seems valid to me regardless of whether the group $a$ and $b$ belong to is finite. Why does this group have to be finite? Where in the proof does it go wrong otherwise?
