The Proof

Before 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_p^k, 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_q^m. Hmm! Clearly, what we want to do is take each group Z_p^k and find a group U_q^m to contain it.

Now, for q ≠ 2, we know from the previous essay that U_q^m is just the cyclic group Z_{(q−1)q^m−1}, which decomposes into the cyclic groups Z_q−1 and Z_q^m−1. (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_q^m that contains Z_p^k.

However, there's a slight complication. Although the same prime p can appear in any number of the Z_p^k, 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_q^m−1 to take care of one group Z_p^k, most of the time we will need to use the factor Z_q−1 instead.

Now, since q−1 isn't prime (unless q = 3), the group Z_q−1 will decompose further; what we need is for one of the factors to be the group Z_p^k … that is, for p^k to be a divisor of q−1. 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_p^k 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_p^k 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 detail-oriented, here are a few final technical notes.

• When using q = p in the case p = 2, one must in most cases set m = k+2 rather than k+1 due to the factor of Z₂ that drops out.
• It's OK if a prime q happens to be divisible not just by p^k but by some higher power p^r, because the resulting group Z_p^r will contain Z_p^k as a subgroup.
• A single prime q may be suitable for several different values of p and k. For example, q = 53 is suitable for 2, 2², and 13. You just have to be careful not to choose duplicates.
• Actually, the same prime q can be duplicated, as long as it's for different values of p—the different factors don't interfere with one another in that case.