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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.

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.

more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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