Here's the question:

I can do this if <ab> and <ba> are finite. Let the order of <ba> be m. Then form the following product:Quote:

Given a group G and two elements a, b in G, show that |ab| = |ba| where |x| is the order of the subgroup generated by x.

$\displaystyle (ab)^{m+1} = (ab)(ab)....(ab)$

Now, since G is associative we may insert the parenthesis thus:

(ab)(ab)(ab)....(ab)(ab) = a(ba)(ba)(ba)....(ba)b.

But |ba| = m, so

(ab)(ab)(ab)....(ab)(ab) = a(ba)(ba)(ba)....(ba)b = a(e)b = ab

But that means that

$\displaystyle (ab)^{m+1} = ab \implies (ab)^m = e$

which implies that the order of <ab> is m.

What's this have to do with induction? Well, I can use this method to show that |ab| = |ba| for when the orders are countably infinite. My question is what if the orders have a greater cardinality than $\displaystyle \aleph _0$. My thought was to use transfinite induction, but I have never seen it used. I am guessing that I can't...the index set for the problem is $\displaystyle \mathbb{N}$ because of the fact that I am using n as an exponent and thus n is a natural number.

Otherwise I am sure there is a more elegant way of proving this theorem which could possibly solve the problem of infinite orders. Any suggestions?

-Dan