Algebra - Elements of Infinite Order

[m0bius]

Homework Statement

Let G be a group and let a be an element in G. Show that a and a^-1 generate the same cyclic subgroup <a> = <a^-1> and have the same order o(a) = o(a^-1).

Homework Equations

This is in the second section of the book I'm using, so there aren't really any theorems that can be used, just the definition of a group, order, and cyclic groups.

The Attempt at a Solution

First we show o(a) = o(a^-1).

Assume a =/= e, and that both o(a) and o(a^-1) are finite.
Let o(a) = n and o(a^-1) = m. Then aⁿ = e and (a^-1)^m = e by the definition of order. Since a =/= e, m > 1 and n > 1. Multiply these two equations together to get:

(aⁿ)((a^-1)^m) = e. (1)

Case 1 : Assume m > n. Then we have (a^-1)^m-n = e. This can be seen by expanding (1), then canceling the a's with the a^-1's from the inside outwards. Since n > 1, we can say m > 2. But then 0 < m - n < m, and so m is not the least positive integer such that (a^-1)^m = e. So o(a^-1) =/= m and this is a contradiction.

Case 2 : Assume m < n. Then we have a^n-m = e, which can be seen by expanding and canceling like we did in Case 1. Since m > 1, we can say n > 2. But then 0 < n - m < n, and so n is not the least positive integer such that aⁿ = e. So o(a) =/= n and this is a contradiction.

Since m is not greater than n, and n is not greater than m, we must have n = m, and so o(a) = o(a^-1).

I think I made it longer than it needed to be, but it seems correct. Anyway, my problem is now with the case when either or both of o(a) and o(a^-1) are infinite. Let me just say I have about 5 pages or so of frantic scribblings trying to figure out what would happen in this case, and I am now extremely confused. My book only mentions one example at this point of an element with infinite order : all n =/= 0 in {Z, +} have infinite order (I understand this).

I'm not sure if it even makes sense to say that two elements of infinite order have the same order, but I'm currently attempting to show that if o(a) is infinite then o(a^-1) is infinite. Any help/hints would be appreciated.

Thanks,
m0bius.

 

[micromass]

Assume that o(a^-1) is infinite. Proceed by contradiction, thus assume that o(a) is finite. Then your proof in the OP shows that o(a^-1) is finite. Contradiction...

 

[m0bius]

Assume that o(a^-1) is infinite. Proceed by contradiction, thus assume that o(a) is finite. Then your proof in the OP shows that o(a^-1) is finite. Contradiction...

Hey, sorry for the delayed response. I'm not sure if this works though. We assume that o(a^-1) is infinite, and for the proof by contradiction we assume that o(a) is finite, but the proof in the OP requires that both be finite for it to work.

I've been working on this a bit though and I think I've found a proof, although it's basically a bunch of little "theorems". In the OP we proved :

Theorem 1 : Let G be a group and a an element of G. If o(a) and o(a^-1) are both finite, then o(a) = o(a^-1).

Note for everything below we are assuming o(a) > 1, since the case when a = e is trivial to solve and can be handled separately.

Theorem 2 : o(a) is finite if and only if o(a^-1) is finite.

Proof.
>>> First assume o(a) = n and n is finite, so aⁿ = e. The proof is by contradiction, so assume o(a^-1) is infinite. Then (a^-1)^m =/= e for all positive integers m (for all m in Z⁺). Multiply a^-1 to both sides of aⁿ = e to get a^-1 = a^n-1. Then we see:

(a^-1)^m =/= e implies (a^n-1)^m =/= e which implies (a^m(n-1)) =/= e. However since n > 1, m(n-1) is in Z⁺, so o(a) is infinite, which is a contradiction.

<<< Next assume o(a^-1) = m and m is finite. The proof is by contradiction, so assume o(a) is infinite. Then aⁿ =/= e for all n in Z⁺. Using a^-1 = a^n-1, we get a^m(n-1) = e, but since m(n-1) is in Z⁺ this means o(a) is finite, which is a contradiction.

Theorem 3 : o(a) is infinite if and only if o(a^-1) is infinite.

>>> Assume o(a) is infinite. The proof is by contradiction, so assume o(a^-1) is finite. Then by Theorem 2 o(a) is finite, which is a contradiction.

<<< The proof in the other direction is similar.

OK...so for any group G and any element a in G, o(a) = o(a^-1) by theorems 1,2, and 3.

Now I still need to show <a> = <a^-1>...

 

[micromass]

Alright, that works! As for the thing you need to prove, can you show that $$a^{-1}\in <a>$$? This would imply that $$<a^{-1}>\subseteq <a>$$. The other inclusion follows analogously...

 

[Deveno]

if o(a) is finite, then since <a> is a group, a^-1 is in it, right?

(in fact, if o(a) = k, then a^k-1 must be a^-1 (why?))

so a^-1 is in <a>, hence <a^-1> is contained in <a>.

now what can the order of a^-1 possibly be?

we know that for any m, (a^-1)^m = (a^m)^-1 so...

but if o(a) = o(a^-1), how much bigger than <a^-1> can <a> be?

it gets more interesting if o(a) is infinite. here, a^-1 isn't ANY of the positive powers of a. so <a> must consist of all INTEGRAL powers of a (we have to add a^-1 and its powers to all the positive powers of a, as well as the identity.).

so use the fact that (a^-1)^k = a^-k = (a^k)^-1, and that a = (a^-1)^-1