Is it true that $O(ab)=O(ba)$, where $G$ is a group and $a,b \in G$? 
Is it true that $O(ab)=O(ba)$, where $G$ is a group and $a,b \in G$?

Suppose $O(a)$ and $O(b)$ are finite and also $O(ab)$ and $O(ba)$ are finite. Then $\operatorname{lcm} (|a|,|b|) = \operatorname{lcm}(|b|,|a|)$. (Is that correct?)
Suppose $O(a)$ and $O(b)$ is finite but $O(ab)$ is infinite, how to prove that $O(ab) = O(ba)$? 

If not give elements $a,b$ in a group $G$, such that $O(ab)$ is infinite but $O(ba)$ is finite.

 A: Just to be clear, the order of $ab$ isn't necessarily the LCM of the orders of $a$ and $b$.  For example, consider the cycles $a=(1,2)$ and $b=(1,2,3)$ in $S_3$: the LCM of their orders is 6, but $ab=(1,3)$ (or $(2,3)$ if right-to-left) has order 2.
However order is invariant under isomorphisms.  More precisely, if $\phi: G \rightarrow H$ is a group isomorphism and $g \in G$, the order $|\langle g \rangle|$ of $g$ in $G$ is the same as $|\langle \phi(g) \rangle|$, the order of $\phi(g)$ in $H$.
In particular, conjugation by $b$ is an automorphism of $G$: define $\phi: G \rightarrow G$ by  $\phi(g) = b^{-1} g b$. Then $ab = \phi(ba)$ so the two have the same order (whether finite or infinite).
A: (1) Conjugate elements have the same order, i.e. $\;\mathcal o(a)=\mathcal o(g^{-1}ag)\;$
(2) $\;ab=b^{-1}(ba)b\;$
A: Lemma: If the order of ab is infinite, then the order of ba is infinite.
Proof. Contrary to our claim, let $o(ba)= m$. So, $\underbrace{bababa\cdots ba}_{m}=e$. then, $\underbrace{abab\cdots ab}_{m-1}= b^{-1}a^{-1}=(ab)^{-1}$. Hence, $o(ab)|m$, which is a contrast. 
So, you don,t need to have the orders of $a$ and $b$ are finite.
A: Suppose $(ab)^k=1$. Then $(ab)^k aa^{-1} = a (ba)^k a^{-1}=1$.  Thus, $(ba)^k=1$.  
A: If $a$, $b$ commute, and the orders of $a$, $b$ are relatively prime, then the order of $ab$ is $O(ab)=O(a)\cdot O(b)$. 
