Only Two Abelian Groups
Date: 02/25/2003 at 18:56:08 From: Anonymous Subject: College Algebra Show that any group with order p^2, where p is a prime, is Abelian. Show that up to isomorphism that only two such groups exist. Let G be a group with C the center of G. C is a subgroup of G. By Lagrange's theorem |C| divides |G|. Thus |C|=1,p,p^2. If |C|=p^2, then C=G and thus G is Abelian. However I am stumped when it comes to ruling out |C|=1 and |C|=p and I have no clue how to find the two groups that are supposed to exist. Any finite group G of prime power order has center C/=1. The proof is unclear to me. Therefore I am not going to rule out |C|=1 just yet. Even so, I still have the case of |C|=p. Any thoughts will help!
Date: 02/26/2003 at 04:09:19 From: Doctor Jacques Subject: Re: College Algebra Hi, The crucial point is indeed that the center of a group of prime power order is not trivial. This is usually proven by using the concepts of group action on a set and conjugation. Let us assume for the moment that we know that |C| <> 1, which implies that |C| = p. As the center is a normal subgroup, it defines a quotient group G/C, which is of order p, and therefore cyclic. Let gC be a generator of that cyclic quotient group. Every element x of G can be written as (g^i)*c, for some c in C and some integer i. Consider two elements x and y of G. We have: x = (g^i)*c_1 y = (g^j)*c_2 with c_1 and c_2 in C. This yields: xy = (g^i)*c_1*(g^j)*c_2 = (g^(i+j))*c_1*c_2 because c_1 and c_2 are in C. If you compute yx in the same way, you will find that yx = xy, i.e. any two elements commute, and C = G after all. This shows that G is abelian; in fact any group G with center C such that G/C is cyclic is abelian. There are two cases to consider: Case 1 ------ G has an element of order p^2. In this case, G is cyclic of order p^2 Case 2 ------ G has no element of order p^2. In this case, all non-identity elements of G are of order p. Consider one such element a, and let H = <a> be the cyclic group (of order p) generated by a. Let b be an element not in H, and K = <b> be the cyclic subgroup (of order p) generated by b. We claim that G is isomorphic to H x K (the direct product). As G is abelian, we need only to prove: (1) (H intersect K) is the trivial subgroup. (2) Every element of G can be written as hk, with h in H and k in K. To prove (1), assume that x <> e belongs to H and K. As every non-identity element of a cyclic group of prime order is a generator, this means that x is a generator of both H and K, and that these subgroups are identical, which contradicts the choice of b. To prove (2), consider the quotient G/H (as G is Abelian, we are free to take quotients). G/H being of order p, it is cyclic. As b is not in H, bH is a generator of G/H. We can therefore write any element x of G as: x = (b^i)*h with h in H As b^i is in K, this proves (2). To summarize, G can only be cyclic of order p^2, or isomorphic to the direct product C_p x C_p, where C_p is the cyclic group of order p. Does this help? Please feel free to write back if you want to discuss this further. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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