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Group Proof

Date: 02/22/2002 at 00:33:43
From: Bethany
Subject: Modern algebra

Hi. I have to:

Prove that if G is a group of order p^2 (p is prime), and G is not 
cyclic, then a^p = e for each a in G.  

The only thing I can think of is using p = mq+r, then saying 
a^p = a^(mq+r), then ((a^m)^q)*a^r, then somehow saying that a^p = e.

Thanks for your help,

Date: 02/22/2002 at 23:08:11
From: Doctor Paul
Subject: Re: Modern algebra

As an easy corollary of Lagrange's Theorem, we know that if x is an 
element of G, then the order of x must divide the order of the group 
(to see this, just notice that the order of x is just the order of the 
subgroup generated by x).

Since |G| = p^2, |x| = 1, p, or p^2 for all x in G.

Clearly, |x| is not equal to p^2 for any x in G because then G would 
be cyclic.

Thus we either have |x| = 1 or |x| = p.

The only element satisfying |x| = 1 is the identity element and it is 
certainly true that the identity element raised to the pth power is 
again the identity (ie, e^p = e).

The only possibility for every other nonidentity element is that it 
must satisfy |x| = p.  This completes the proof.

I hope this helps.  Please write back if you'd like to talk about this 

- Doctor Paul, The Math Forum   
Associated Topics:
College Modern Algebra

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