if $a,b \in G$ and $ab$ has finite order n, then $ba$ has order n

Question

if $a,b \in G$ and $ab$ has finite order $n$, then $ba$ has order n

Okay this question is solved on: Show that if $ab$ has finite order $n$, then $ba$ also has order $n$. - Fraleigh p. 47 6.46.

Now my question would be. What is meant by $ab$ has finite order $n$? I'm trying to understand the notation here. $ab$ doesnt seem like a group to me, it seems like just some multiplication which yields another element of $G$. How did we assume $ab$ is a cyclic group?

Second question, assuming $ab$ is a cyclic group makes sense but what is only given to us is that ab has finite elements. do only cyclic groups have finite elements? I don't think so Concluding that $(ab)^n=e$ would help us a lot but I can't conclude that directly.

Answer 1

You seem to be unaware that the word order is applied to elements as well as groups. The order of an element is the order of the cyclic subgroup it generates.

$a,b, ab$ and $ba$ are all elements of the group, not groups themselves.

Terminology tends to get overloaded in mathematics, so you always have to be sensitive to the context to detect issues like these.

Answer 2

The order of an element $x$ of a group is the smallest $n$ such that $x^n$ is the identity. This can indeed be interpreted as the order of a group, but not necessarily the order of the whole group $G$: it is the order of the cyclic subgroup consisting of all powers of $x$.