Group Theory: ord(ab) = ord(ba) - Proof If for some group with a binary operation, (G, *), we have ord $a$ := $a^n = 1$ for some positive integer $n$, would it be correct to argue that:
Case 1 (finite order): ord $ab$ = ord $ba$:
$(ab)^n = a^n b^n = 1 \iff b^n a^n b^n = b^n \iff b^n a^n (b^n (b^{-1})^n)=b^n (b^{-1})^n \iff $ $b^n a^n = (ba)^n= 1$
Case 2 (infinite order): ord $ab$ = ord $ba$:
ord $ab \to \infty$ as $n \to \infty$, and therefore ord $ab$ = ord $ba$
Do you think my proof is reliable?
 A: In general, $(ab)^n = a^n b^n$ is not true in groups.
For a finite order, I advise you to write
$$(ab)^n = abababab\dots ab = a(bababa\dots ba)a^{-1}$$
A: Let $G$ be a group. The element order of $g\in G$ is defined as the group order of the cyclic subgroup
$\langle g\rangle=\{g^n|n\in\mathbb Z\}$.
We have $b\langle ab\rangle=\{b(ab)^n|n\in\mathbb Z\}=\{(ba)^nb|n\in\mathbb Z\}=\langle ba\rangle b$.
Remember that for a subgroup $H\subseteq G$, all cosets $gH$ and $Hg$ have the same order as $H$.
We conclude $\operatorname{ord}(ab)=|\langle ab\rangle|=|b\langle ab\rangle|=|\langle ba\rangle b|=|\langle ba\rangle|=ord(ba)$.
(Note that with this proof, we need not distinguish between the finite and the infinite case.)

Alternatively, if you don't like using cosets, there's another easy proof for both cases at once:
This proof is more suitable if your lecture uses the definition $\operatorname{ord}(g)=\min\{n\in\mathbb N|g^n=1\}$.
$ab$ and $ba$ are conjugates, since $a^{-1}(ab)a=ba$.
Consequently, we have $(ba)^n=(a^{-1}(ab)a)^n=a^{-1}(ab)^na\ \forall\ n\in\mathbb Z$.
We conclude $(ba)^n=1\ \Leftrightarrow\ a^{-1}(ab)^na=1\ \Leftrightarrow\ a(a^{-1}(ab)^na)a^{-1}=aa^{-1}\ \Leftrightarrow\ (ab)^n=1$.
Therefore $\operatorname{ord}(ab)$ and $\operatorname{ord}(ba)$ must coincide.
