Modern Algebra Proof

Date: 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/