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Eigenray
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Elements of finite order
« on: Jan 17th, 2005, 8:02pm » |
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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.
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Icarus
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Re: Elements of finite order
« Reply #1 on: Jan 20th, 2005, 6:25pm » |
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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.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Eigenray
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Re: Elements of finite order
« Reply #2 on: Jan 25th, 2005, 3:47pm » |
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You're on the right track, noting that F is stable under taking inverses and conjugation.
Perhaps I should give a hint?
Consider <F>, the subgroup generated by F. Obviously you want to show <F>=F. An indirect method is to note that it suffices to show <F> is finite.
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Icarus
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Re: Elements of finite order
« Reply #3 on: Jan 25th, 2005, 8:07pm » |
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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.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
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Re: Elements of finite order
« Reply #4 on: Feb 14th, 2005, 8:18pm » |
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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.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Eigenray
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Re: Elements of finite order
« Reply #5 on: Feb 16th, 2005, 11:51am » |
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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.
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Eigenray
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Re: Elements of finite order
« Reply #6 on: Mar 15th, 2005, 5:31pm » |
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Okay, a more explicit hint:
Show there is some N such that Fn=FN for all n>N.
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| « Last Edit: Mar 15th, 2005, 5:36pm by Eigenray » |
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Eigenray
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Re: Elements of finite order
« Reply #7 on: Aug 15th, 2005, 1:38pm » |
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I don't think I can give any more hints without giving it away completely...
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astrix
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Re: Elements of finite order
« Reply #8 on: Nov 26th, 2005, 9:44am » |
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Icarus, what is [bbz] in that set?
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Icarus
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Re: Elements of finite order
« Reply #9 on: Nov 26th, 2005, 11:48am » |
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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 (]).
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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