# Transfinite Induction?

• Mar 31st 2011, 02:02 PM
topsquark
Transfinite Induction?
Here's the question:
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.
I can do this if <ab> and <ba> are finite. Let the order of <ba> be m. Then form the following product:
$(ab)^{m+1} = (ab)(ab)....(ab)$

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
$(ab)^{m+1} = ab \implies (ab)^m = e$
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 $\aleph _0$. 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 $\mathbb{N}$ 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
• Mar 31st 2011, 02:31 PM
tonio
Quote:

Originally Posted by topsquark
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:
$(ab)^{m+1} = (ab)(ab)....(ab)$

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
$(ab)^{m+1} = ab \implies (ab)^m = e$
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 Aleph _0. (I can't find the LaTeX code for Aleph.) 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 $\mathbb{N}$ 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

1) First show that two conjugate elements in a group have the same order, i.e.

$x=g^{-1}yg\Longrightarrow ord(x)=ord(y)$

2) Now show that $ab\,,\,\,ba$ are always conjugate in any group...

Tonio
• Mar 31st 2011, 03:09 PM
topsquark
Quote:

Originally Posted by tonio
1) First show that two conjugate elements in a group have the same order, i.e.

$x=g^{-1}yg\Longrightarrow ord(x)=ord(y)$

2) Now show that $ab\,,\,\,ba$ are always conjugate in any group...

Tonio

Works for me. Thank you.

-Dan
• Mar 31st 2011, 07:41 PM
Deveno
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:

e <-->1
g <-->2
g^-1 <--> 3
g^2 <--> 4
g^-2 <--> 5 etc. this is just the standard bijection of Z with N+).
• Mar 31st 2011, 09:11 PM
Drexel28
Quote:

Originally Posted by Deveno

(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:

Really? You've never heard of ordinal numbers? Namely, as the title suggests if one can prove that a property is true for all ordinals $\beta<\alpha$ then the property is true for the ordinal $\alpha$ then the property is true for all ordinals. That said, that is really, really heavy machinery to be using here.
• Mar 31st 2011, 09:34 PM
Deveno
Quote:

Originally Posted by Drexel28
Really? You've never heard of ordinal numbers? Namely, as the title suggests if one can prove that a property is true for all ordinals $\beta<\alpha$ then the property is true for the ordinal $\alpha$ then the property is true for all ordinals. That said, that is really, really heavy machinery to be using here.

i mean as the order of the element of a group, not the order of a SET. for g in a group G, <g> is cyclic, and all cyclic groups have countable order (as in the cardinality of the underlying set).

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.