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, Bethany
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 more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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