Modern Algebra ProofDate: 01/29/2001 at 22:55:49 From: Paul Subject: Modern algebra Here's the problem I've been asked to do: Let G be a finite group and suppose 2 divides the order of G. That is, |G| is even. Show that there exists x in G such that x^2 = e (G has an element of order two). Here's what I've done so far: I began by stating a little Lemma and proving it: Lemma: Let a be an element of G. Then a^|G| = e Proof: By a Corollary of Lagrange's Thm, |a| divides |G|. Hence, |G| = |a| * k for some k in N. Then a^|G| = a^(|a| * k) = (a^|a|)^k = e^k = e Now suppose |G| = 2n and x is an element of G. (*) Then by the lemma, x^(2n) = e. It follows that (x^n)^2 = e and so x^n is my element of order two. My proof is flawed in the following way: sometimes, x^n = e and so x^n has order one. Here's an example: Consider the group Z_8 (integers mod 8) with respect to addition. I made the following table: x 0 1 2 3 4 5 6 7 x^4 0 4 0 4 0 4 0 4 So when x is an even element, my proof breaks down. For example, if x=6, then x^n = 6^4 = 6*4 = 24 = 0 mod 8 So it seems that the members of the group that are relatively prime to eight (that is, the members which generate G) are okay candidates. I think my proof breaks down if the chosen element x does not generate the entire group G. Am I right? I tried it with another group: U(14) = {1,3,5,9,11,13} where the operation is multiplication mod 14. The group U(n) is the set of elements less than n that are relatively prime to n, and the operation is multiplication mod n. I completed the table again: x 1 3 5 9 11 13 x^3 1 13 13 1 1 13 Here I have problems with the elements x = 1, x = 9, x = 11 and these are the elements that do not generate the group U(14). I guess I've pretty much got it figured out and I think I know what is causing my problems. But I don't know how to modify my proof. Can I go back to the statement (*) and suppose that x is an element of G with the property that <x> = G instead of just choosing an arbitrary x in G? If so, how can I be sure that such an x exists? For example, in the Klein-four group (which is a group of order 2n), every element squared is the identity so no element generates the group... Any help you can provide would be appreciated. Date: 01/29/2001 at 23:10:32 From: Doctor Rob Subject: Re: Modern algebra Thanks for writing to Ask Dr. Math, Paul. Your work above is commendable, but it won't lead directly to a proof. One way to proceed is this. Every element of x of G has an inverse x^(-1). The sets {x,x^(-1)} partition the group into disjoint sets. Most of the sets have size 2. Throw out all these sets from G. Since |G| is even, this leaves an even number of sets of one element each, in which that element is its own inverse, and hence has order at most 2: x = x^(-1) ==> x^2 = e ==> |x| divides 2 ==> |x| = 2 or 1. Thus there is an even number of elements of orders 2 or 1 combined. There is, of course, exactly one element of order 1, namely the identity e itself. That leaves an odd number of elements of order exactly 2. An odd number is necessarily >= 1, so there is at least one element of order exactly 2. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/