$a,b$ in $G$ has finite order. Then is the order of $ab, ba, a^{-1}b^{-1}$ with conditions finite? Given that two elements $a,b$ in a group have finite order, are the following true:


*

*$ab = ba \implies ab$ has finite order.

*$ab$ has finite order $\implies ba$ has finite order.

*$ab$ has finite order $\implies a^{-1}b^{-1}$ has finite order.


I am not sure where to start. Should I be trying to manipulate the entity on the right using the assumption on the left of each statement until some power of it equals the identity?
 A: You idea is a good way to approach things.
Let's look at each of the cases in turn. You know that $a$ and $b$ have finite order, so there exist some $m,n\in\mathbb{N}$ such that $a^n=b^m=e$ (where we are assuminng the group is written multiplicatively and that it has identity element $e$).
1) If you assume that $ab=ba$, then what do you know about $(ab)^2=abab$.? Can you find a way of writing this in the form $a^rb^s$ for some $r,s$.? Can you do something similar for $(ab)^i$ for an arbitrary value of $i$.? You should be able to, and then you can use your assumption that $a^n=b^m=e$ to find an appropriate $i$ such that $(ab)^{i}=e$.
2) If $ab$ has finite order, then you know that there exists some $j$ such that $(ab)^j=e$. Can you use this to determine anything about $(ba)^{j+1}$. Try working this out and writing it in the form $b^ra^s$ for some $r,s$. You should then be able to prove part (2).
3) Assuming you have proved part (2), what do you know about $(ba)^{-1}$. In particular, what do you know about it if $(ab)$ (and hence by part (2) $(ba)$) has finite order?
A: All true.


*

*Lets say $o(a)=n,o(b)=m$. If $ab=ba$ then $(ab)^k=abab\ldots ab=a^kb^k$. Is there any $k \in \mathbb{N}$ s.t. $a^k=1 \ \text{and} \ b^k=1$? If yes the order of $ab$ is finite.

*Here observe that $(ba)^k=baba\ldots ba=b(ab)(ab)\ldots (ab)a=b(ab)^{k-1}a$ so if $o(ab)=r$ then $(ba)^{r+1}=ba \Rightarrow (ba)^r=1$.

*Observe that $a^{-1}b^{-1}=(ba)^{-1}$ so using 2. if $(ab)^r=1 \Rightarrow (ba)^r=1 \Rightarrow ((ba)^{-1})^r = ((ba)^r)^{-1}=1$.

A: Yes, your strategy is a good start and since it seems that you are only looking for a hint I'll keep it short. For instance, given that $ab=ba$ it follows that $(ab)(ab)=a(ba)b=a(ab)b=a^2b^2$. More generally (prove by induction), $(ab)^n=a^nb^n$. So, if $k$ is the order of $a$ and $r$ is the order of $b$ then $(ab)^{(kr)}=e$ and thus $ab$ has finite order. Similarly you can approach the other two statements. 
A: The latter two questions are a specific case of something more general, called conjugacy.
Two elements $g, h\in G$ are conjugate if there exists some $k\in G$ such that $k^{-1}gk=h$. Note that that "being conjugate" is an equivalence relation, so we will write $g\sim h$.
I will write $o(g)$ for the order of the element $g\in G$, and I will write $g^k:=k^{-1}gk$ (which is standard, up to exchanging $k$ and $k^{-1}$).
Lemma: If $h=g^k$ then $o(h)=o(g)$.
Proof: This follows immediately from the fact that $h^n=(g^k)^n=(g^n)^k$.
We can now use this lemma to prove points (2) and (3), as follows.


*

*For point (2), $(ab)^a=ba$ so you can use the Lemma immediately.

*For point (3), $(a^{-1}b^{-1})^{a^{-1}}=b^{-1}a^{-1}=(ab)^{-1}$ so $a^{-1}b^{-1}\sim (ab)^{-1}$. As $o(g)=o(g^{-1})$, you are done.
