Showing ab=ba finite order n

I'm not quite sure what this question is asking.

**Problem: ** Let a and b be elements of a group G. Show that if ab has finite order n, then ba also has order n.

It is my understanding that the order of *group* is simply the number of elements in the group, and the order of a *generator* is the number of elements in the set it generates. But... what is the order of any old *element* in G, like the question asks?

Re: Showing ab=ba finite order n

Quote:

Originally Posted by

**tangibleLime** I'm not quite sure what this question is asking.

**Problem: ** Let a and b be elements of a group G. Show that if ab has finite order n, then ba also has order n.

It is my understanding that the order of *group* is simply the number of elements in the group, and the order of a *generator* is the number of elements in the set it generates. But... what is the order of any old *element* in G, like the question asks?

By definition the order of $\displaystyle a\in G$ is the 'group order' of $\displaystyle \langle a\rangle$.

Re: Showing ab=ba finite order n

there are two equivalent defintions for order. one is: the order of the element a, |a|, is the cardinality of the underlying set of the group generated by a, |a| = |<a>|.

but that is a "non-constructive" definition. a more explicit definition, is the order of a, |a| is the smallest positive integer m such that a^m = e.

to see that these two are equivalent: note that if m is the smallest positive integer for which a^m = e, {e,a,a^2,....,a^(m-1)} are all distinct, and this set is <a>

(the tricky part is showing a^-1 is one of these postive powers).

note that it may be the case that a is of infinite order, this is the case for the number 1 in the integers.

EDIT: to attack this problem, note that if n is the order of ab, (ab)^n = e.

now (ab)^n = (ab)(ab)........(ab) (n times)

consider that (ba)^n = (ba)(ba)....(ba) (n times)

= b(ab)^(n-1)a.

now multiply on the left by e = (a^-1)a.

Re: Showing ab=ba finite order n

Thanks for the reply! It cleared things up a lot.

I was just wondering, is $\displaystyle ab$ in this case actually multiplication, or is it another one of those times in abstract algebra when it actually means something completely different? Because if it *is* multiplication, can't I just say that $\displaystyle ab=ba$ since multiplication is commutative and use the order formula $\displaystyle \frac{n}{gdc(n,m)}$ which calculates the number of elements generated by a generator?

EDIT:

Er actually, nevermind. Multiplication isn't commutative for all sets... such as matrices.

Re: Showing ab=ba finite order n

"ab" is written *multiplicatively* when the implied operation is clear from the context.

One definition of a group is a set G, together with an operation (say, *), such that... (axioms).

You can call it "multiplication", but it could mean something very different when you aren't in a commutative (i.e. abelian) group.

Re: Showing ab=ba finite order n

for a group, "ab" is just a notation meaning a(group operation)b. if the operation of (G,*) needs to be emphasized, a*b is often written instead.

this is the hardest thing for most people to get used to in groups, that ab and ba are two different things (perhaps. in abelian groups, they aren't....but unfortunately many groups are not abelian. i know, right?).