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Topic: Abelian Groups (Revisited) (Read 759 times) |
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Barukh
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Abelian Groups (Revisited)
« on: Sep 22nd, 2004, 6:00am » |
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Prove that every group of order p2 (where p is a prime number) is abelian.
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Obob
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Re: Abelian Groups (Revisited)
« Reply #1 on: Oct 20th, 2004, 1:47pm » |
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Suppose that G is not abelian.
Denote by Z(G) the center of G, i.e. the set of all g in G such that gh=hg for every h in G. If |Z(G)|=|G|=p^2, then G must be abelian. Hence, by Lagrange's Theorem, since Z(G) is a subgroup of G it follows that |Z(G)| is either 1 or p.
We claim that |Z(G)| cannot be 1. Consider the action of G on itself by conjugation. The orbits of order one are exactly the elements of Z(G), so #(orbits of order 1)=|Z(G)|. Denote by C_G(g) the set of all elements which commute with g. Now C_G(g) is a subgroup of G, and hence the index [G:C_G(g)] divides G. But it is easy to see that #(conjugates of g)=[G:C_G(g)], so the size of each conjugacy class divides the order of the group. Hence we have |G|=|Z(G)|+[G:C_G(g_1)]+...+[G:C_G(g_k)], where g_1,...,g_k is a complete set of representatives of the orbits with size > 1. Hence each of the terms [G:C_G(g_i)] is either p or p^2, and |Z(G)| must therefore be either p or p^2 as well.
Finally, we claim that if G/Z(G) is cyclic then G is abelian. Set Z=Z(G), and let g be such that G/Z=<gZ>. Now for every x in G, we can find a positive integer k and a z in Z(G) for which x=g^k z. Say y=g^j z'. Then since z and z' are in the center, we have xy=g^k z g^j z'=g^j z' g^k z=yx. Thus if G/Z is cyclic then G is abelian.
Now if |Z(G)|=p then |G/Z(G)|=p. Since all groups of order p are cyclic, G/Z(G) is cyclic, and G is abelian.
This completes the proof.
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