Product of a Finite Abelian GroupDate: 06/28/2004 at 11:19:21 From: Angie Subject: What can I say about the "product" of a finite Abelian group Suppose G = {a1, a2, ... , an} is a finite Abelian group. If G has odd order, what can you say about the "product," a1*a2*...*an, of all the elements of G? What can you say about this "product" if G has even order? What if G is not Abelian? I am not sure how * not being commutative affects the "product" of the elements of the group. My thoughts: If G has odd order, then it has an odd number of elements one of which is the identity and the rest would be elements and their inverses. e*(a1*a2*a3*...*an*a1'*a2'*a3'*...*an')---> e*(a1*a1'*a2*a2'*...*an*an')---> e*e = e But what if some of the elements are their own inverses? e*(a1*a2*a3*a4*...*an*a3'*a4'*...*an') ---> e*(a1*a2)*(a3*a3'*a4*a4'*...*an*an')---> e*(a1*a2)*e ---> e*a1*a2 ---> e*a ---> a where a1*a2 = a, which would be an element of the group G (G is closed under *). If G has even order, then it has an even number of elements one of which is the identity, e, and then an odd number of elements. There would have to be an odd number of elements that are there own inverses and an even number of elements with an even number of inverses? e*(a1*a2*a3*a4*a5*...*an*a4'*a5'*...*an') ---> e*(a1*a2*a3*a4*a4'*a5*a5'*...*an*an') ---> e*(a1*a2*a3)*(a4*a4'*a5*a5'*...*an*an') ---> e*(a1*a2*a3)*e ---> e*(a1*a2*a3) ---> e*a ---> a where a1*a2*a3 = a and a is an element of G since G is closed under *. Date: 06/28/2004 at 20:51:20 From: Doctor Vogler Subject: Re: What can I say about the "product" of a finite Abelian group Hi Angie, We got a question very similar to yours a short time ago, so I've thought about this before. It sounds like you might be forgetting that your group is not necessarily cyclic. But it is close. The Fundamental Theorem of Finitely Generated Abelian Groups says that all finite Abelian groups are a product of cyclic groups. Use that fact, and then it might help to write it out additively instead of multiplicatively. Suppose that our group is (Z/aZ) x (Z/bZ) x (Z/cZ). How do we find the sum of all of the a*b*c elements of this group? Remember that we add termwise, so we have a-1 b-1 c-1 a-1 b-1 c-1 sum sum sum (r, s, t) = ( b*c*sum r, a*c*sum s, a*b*sum t ) r=0 s=0 t=0 r=0 s=0 t=0 = (b*c*a*(a-1)/2, a*c*b*(b-1)/2, a*b*c*(c-1)/2). Now the first term is considered mod a, the second mod b, the third mod c. At first glance, you think: "a divides the first term, b the second, and c the third, so the sum is the identity (0, 0, 0)," but that's not quite true; you have to divide by 2 first. If a, b, or c is odd, then 2 has an inverse, and so the corresponding term *is* zero. If not, then you have to divide one of the even numbers in the product by 2 first. Suppose that two of a, b, and c are even. Then each term will still be zero, because you can divide the two out of the other number. But if only one of a, b, or c is even (say a), then you get (b*c*(a/2)*(a-1), (a/2)*c*b*(b-1), (a/2)*b*c*(c-1)) = (b*c*(a/2)*(a-1), 0, 0) since the last two terms are divisible by b and c, respectively. Furthermore, (a/2)*k mod a is determined by whether k is odd or even, since 2*(a/2) = 0 mod a, and since b, c, and a-1 are all odd, we have (a/2, 0, 0). Since a/2 is the unique element of order 2 in Z/aZ, and the other component groups have odd order, this is the unique element of order 2 in our Abelian group, that is, the element -1 when the group is considered multiplicatively. I used three cyclic components to have a concrete example to use, but the argument does not depend on the three. You can generalize this to any finite Abelian group, and the result is the same: If there is exactly one cyclic component of even order, then the sum (product) of all of the elements is the unique element of order 2. Otherwise, the sum (product) of all of the elements is the identity. In particular, the cyclic group of order 8 is different from the product of three cyclic groups of order 2 (which group also has order 8). In the second group, the sum (product) of all 8 elements is the identity, but not in the first. For a concrete example, let us examine the product of all elements less than n and relatively prime to n in the finite Abelian group (Z/nZ)*, the multiplicative group of invertible elements mod n. Another well-known theorem from abstract algebra tells exactly how this group is structured: If n = r*s and r and s are relatively prime, then we have the isomorphism (Z/nZ)* = (Z/rZ)* x (Z/sZ)*. If n = p^k for a prime p and k>=1, then we have the isomorphism (Z/nZ)* = (Z/(p-1)Z) x (Z/(p^(k-1))Z) for odd p, (Z/nZ)* = (Z/2Z) x (Z/(2^(k-2))Z) for p=2 and k>1, and (Z/2Z)* = 1 (the trivial group). So you can use this to decompose any group. For example, (Z/54Z)* = (Z/2Z)* x (Z/27Z)* = 1 x (Z/2Z) x (Z/9Z) = (Z/2Z) x (Z/9Z), which has one component of even order, so the product of the elements is -1. In fact, using all of these ideas, it is not hard to show that the product of the elements of (Z/nZ)* is -1 if and only if n is: 2 (in which case 1 = -1), 4, p^k (for some odd prime p), or 2*p^k (again, for an odd prime p). In the case n = 2 and all other cases, the product is 1. Finally, you asked about non-Abelian groups. For this, I might ask you exactly what you mean by the product of all of the elements. For example, in the non-Abelian group S3, of order 6, you might mean 1 (12) (23) (13) (123) (132) or you might mean 1 (123) (12) (13) (132) (23) or any of a variety of other things. Do you always get the same product? Do you when the group is Abelian? If you have any questions or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 06/28/2004 at 21:00:00 From: Angie Subject: Thank you (What can I say about the "product" of a finite Abelian group) Thank you, Dr. Vogler. I think that the "product" would be the same no matter what the order of the elements. I don't understand how not being able to change the order in which the operation acts on the elements would make any difference in the final product. Date: 06/29/2004 at 17:13:42 From: Doctor Vogler Subject: Re: What can I say about the "product" of a finite Abelian group Hi Angie, In my noncommutative example, I mentioned 1 (12) (23) (13) (123) (132) and 1 (123) (12) (13) (132) (23). The first product is (32). The second is (13). The order changes the value of the product. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 06/29/2004 at 18:14:40 From: Angie Subject: Thank you (What can I say about the "product" of a finite Abelian group) Thank you. I have to remember that the "operation" is not just multiplication. Thanks again. |
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