# $a,b$ 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:

1. $ab = ba \implies ab$ has finite order.
2. $ab$ has finite order $\implies ba$ has finite order.
3. $ab$ 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?

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For (1) just compute $(ab)^n$ in terms of $a$ and $b$, for (2) note that $(ab)^n = abab\cdots ab = a(ba)^{n-1}b$ for (3) use (2) and $a^{-1}b^{-1} = (ba)^{-1}$ –  martini Oct 31 '12 at 9:06

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?

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All true.

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

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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 $k^{-1}gk=h$. We often write $g\sim h$ for the equivalence relation, and $g^k:=k^{-1}gk$.

Note that that "begin conjugate" is an equivalence relation.

Lemma: If $h=g^k$ then $o(h)=o(g)$.

Proof: This follows immediately from the fact that $h^n=(g^k)^n=(g^n)^k$.

So, 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.

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The notation $g^k$, where $g,k\in G$, is very well chosen. –  YACP Apr 22 at 20:02