Home

> urticator.net
Search

> 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 Un
 > The Proof
An Example

## The Proof

prev

Before getting into the details, let me restate the theorem.

Every finite abelian group is isomorphic to a subgroup of Un 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 Zpk, and we're trying to fit it into some group Un, which we can write as a product of groups of the form Uqm. Hmm! Clearly, what we want to do is take each group Zpk and find a group Uqm to contain it.

Now, for q != 2, we know from the previous essay that Uqm is just the cyclic group Z(q-1)qm-1, which decomposes into the cyclic groups Zq-1 and Zqm-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 Uqm that contains Zpk.

However, there's a slight complication. Although the same prime p can appear in any number of the Zpk, 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 Zqm-1 to take care of one group Zpk, most of the time we will need to use the factor Zq-1 instead.

Now, since q-1 isn't prime (unless q = 3), the group Zq-1 will decompose further; what we need is for one of the factors to be the group Zpk … that is, for pk to be a divisor of q-1. In other words, we need to be able to find a prime q of the form

q = pk j + 1.

In fact, since there can be any number of groups Zpk 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 Zpk 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 pk but by some higher power pr, because the resulting group Zpr will contain Zpk 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.

next