# Showing ab=ba finite order n

• Oct 3rd 2011, 08:28 PM
tangibleLime
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?
• Oct 3rd 2011, 08:38 PM
Drexel28
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 $a\in G$ is the 'group order' of $\langle a\rangle$.
• Oct 4th 2011, 01:28 AM
Deveno
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.
• Oct 4th 2011, 07:35 PM
tangibleLime
Re: Showing ab=ba finite order n
Thanks for the reply! It cleared things up a lot.

I was just wondering, is $ab$ 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 $ab=ba$ since multiplication is commutative and use the order formula $\frac{n}{gdc(n,m)}$ which calculates the number of elements generated by a generator?

EDIT:
Er actually, nevermind. Multiplication isn't commutative for all sets... such as matrices.
• Oct 4th 2011, 07:42 PM
TheChaz
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.
• Oct 4th 2011, 11:17 PM
Deveno
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?).