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: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.
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
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 . 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 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?
another possibility: suppose that |ab| is infinite (we don't really care "how infinite").
that means that (ab)^n ≠ e, for any positive integer n.
but if |ba| is finite, then ba = (ba)(e) = (ba)^(1+|ba|) = b[(ab)^(|ba|)]a, and hence (ab)^(|ba|) = e, contradiction.
therefore, |ab| infinite implies |ba| infinite.
(as far as countability and order goes, there is no notion that i'm aware of that distinguishes between "countably infinite order" and "uncountably infinite order". but even if there were, the cardinality of <g> is always countable:
g^-1 <--> 3
g^2 <--> 4
g^-2 <--> 5 etc. this is just the standard bijection of Z with N+).
in other words |g| = |<g>|. while as a set (the right-hand side), it's conceivable that |<g>| could be uncountable (for example the additive group of real numbers is an example), it turns out that this never happens (all infinite cyclic groups are (group) isomorphic to Z, which is countable, and finite cyclic groups are finite). so the left-hand side is NEVER an ordinal greater than, oh what do you experts call it? ω, that's it.
my point is, transfinite induction is not needed, here, orders of group elements fall into 2 categories: finite, and non-finite. it doesn't matter what KIND of non-finite you mean.