Home> urticator.net Search About This Site > Domains Glue Stories Computers Driving Games Humor Law > Math Numbers Science Game Theory Section > A Theorem on Finite Abelian Groups Probability Miscellaneous The Structure of U_{n}

The ProofBefore getting into the details, let me restate the theorem.
Every finite abelian group is isomorphic to a subgroup of U_{n} for some appropriate n. Now, in trying to prove anything about all finite abelian groups, it's very helpful to know the result that my algebra textbook calls the Fundamental Theorem on Finite Abelian Groups.
Every finite abelian group is the direct product of cyclic groups. Not only that, but, as I mentioned at the end of the previous essay, we can decompose the cyclic groups into prime power factors. It's not really necessary to do so, but I like to anyway. So, we have some abelian group G, which we can write as a product of groups of the form Z_{pk}, and we're trying to fit it into some group U_{n}, which we can write as a product of groups of the form U_{qm}. Hmm! Clearly, what we want to do is take each group Z_{pk} and find a group U_{qm} to contain it. Now, for q != 2, we know from the previous essay that U_{qm} is just the cyclic group Z_{(q1)qm1}, which decomposes into the cyclic groups Z_{q1} and Z_{qm1}. (In fact, the first factor usually decomposes further.) So, all we need to do is set q = p and m = k+1, and we're done—we've found a group U_{qm} that contains Z_{pk}. However, there's a slight complication. Although the same prime p can appear in any number of the Z_{pk}, the primes q must be distinct, because they're supposed to be the prime factors of the number n. Thus, although we can use the factor Z_{qm1} to take care of one group Z_{pk}, most of the time we will need to use the factor Z_{q1} instead. Now, since q1 isn't prime (unless q = 3), the group Z_{q1} will decompose further; what we need is for one of the factors to be the group Z_{pk} … that is, for p^{k} to be a divisor of q1. In other words, we need to be able to find a prime q of the form
q = p^{k} j + 1. In fact, since there can be any number of groups Z_{pk} with the same values of p and k, we need to be able to find a whole series of such primes. Fortunately, it is possible to do just that. According to Dirichlet's theorem, given any two numbers a and b that are relatively prime, there are an infinite number of primes of the form
q = aj + b. And there you have it! We can find a suitable, distinct prime q for each factor Z_{pk} of the original group G; and multiplying together those primes, along with any prime powers obtained by setting q = p, yields the desired integer n. For the detailoriented, here are a few final technical notes.

See AlsoMultiplication in Base 10 Theorem on Finite Abelian Groups, A @ May (2001) 