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        riddles >> putnam exam (pure math) >> Elements of finite order
        
(Message started by: Eigenray on Jan 17^th, 2005, 8:02pm)

Title: Elements of finite order
Post by Eigenray on Jan 17^th, 2005, 8:02pm

Let G be a group, and let F be the subset consisting of all elements of finite order. If F is finite, show that it is a subgroup.

Title: Re: Elements of finite order
Post by Icarus on Jan 20^th, 2005, 6:25pm

I've been playing with this one, but have still not found an answer. But I will give what I have thus far and see if anyone else wants to contribute. I believe the key to the whole thing lies in convolutions:

If x, y [in] G, then the convolution x * y of x by y is defined by:
x * y = yxy^-1.

The following facts about convolutions are trivial to prove:
(1) e * x = e and x * e = x
(2) (xy) * z = (x * z)(y * z)
(3) x^-1 * y = (x * y)^-1
(4) (x * y) * z = x * (yz)
(5) xy = yx if and only if x * y = x.

Only slightly less easy is:
(6) For all x, y, o(x * y) = o(x).

Proof: Let n = o(x), then by (2), (x * y)ⁿ = xⁿ * y = e * y = e. Therefore o(x * y) [le] o(x). Also, o(x) = o(x * e) = o(x * y * y^-1) [le] o(x * y). Hence o(x) = o(x * y).

Now, since o(x^-1) = o(x), x^-1 is in F whenever x is. In order to show F to be a subgroup, it is only necessary to show that if x, y [in] F, then xy [in] F also.

Let x, y [in] F. By (6), {x * (xy)ⁿ : n [in] [bbz]} [subseteq] F. Since F is finite, this can consist of only a finite number of distinct elements. Therefore [exists] m, n, with m > n, such that x * (xy)^m = x * (xy)ⁿ. From which it follows that x * (xy)^m-n = x.

Hence there exists N > 0 such that x * (xy)^N = x. In other words, (xy)^Nx = x(xy)^N. Canceling the x's on the left and regrouping gives (yx)^N = (xy)^N.

That is as far as I have gotten.

Title: Re: Elements of finite order
Post by Eigenray on Jan 25^th, 2005, 3:47pm

You're on the right track, noting that F is stable under taking inverses and conjugation.
Perhaps I should give a hint?
Consider [hide]<F>, the subgroup generated by F[/hide]. Obviously you want to show [hide]<F>=F[/hide]. An indirect method is to note that it suffices to show [hide]<F> is finite[/hide].

Title: Re: Elements of finite order
Post by Icarus on Jan 25^th, 2005, 8:07pm

Sorry I've let this one sit. I got distracted by some thoughts on Jock curve covering problem (I'm sure you've seen my painful progress there), and have not gotten back to this one.

Since topology and geometry are more my mainstay, that one has been bothering me for a while.

Title: Re: Elements of finite order
Post by Icarus on Feb 14^th, 2005, 8:18pm

I am still stuck on this problem. It is trivial that if <F> is finite, then <F>=F, but I'm not seeing a way to show that <F> is finite.

Title: Re: Elements of finite order
Post by Eigenray on Feb 16^th, 2005, 11:51am

Since F is closed under inverses,
<F> = F U F² U F³ U ...,
where Fⁿ is the set of all products of n elements from F.

Title: Re: Elements of finite order
Post by Eigenray on Mar 15^th, 2005, 5:31pm

Okay, a more explicit hint:
Show there is some N such that Fⁿ=F^N for all n>N.

Title: Re: Elements of finite order
Post by Eigenray on Aug 15^th, 2005, 1:38pm

I don't think I can give any more hints without giving it away completely...

Title: Re: Elements of finite order
Post by astrix on Nov 26^th, 2005, 9:44am

Icarus, what is [bbz] in that set?

Title: Re: Elements of finite order
Post by Icarus on Nov 26^th, 2005, 11:48am

When I posted that, it was in expectation that William would soon restore the symbolry that we used to have before the YaBB upgrade. [bbz] would display the common symbol for the Integers (Z, in a script font). Some of the others: [subseteq] was the subset sign with the underline (C is the best I can do now), [in] was the element sign (c), [le] was less than or equal to (<), and [exists] was the "there exists" sign (]).