Filed under: Algebra, Math | Tags: algebra, categorical, group theory, MaBloWriMo, Math

Ugh, so, I’ve been really busy today and haven’t had the time to do a Banach-Tarski post. Since I really do want to see MaBloWriMo to the end, I’m going to take a break from the main exposition and quickly introduce something useful. There are a couple major ways of combining two groups into one. The most important one, called the direct product, is analogous to the product of topological spaces. I know this is sort of a wussy post — sorry.

The easiest construction is the direct product. Let be a family of groups indexed by . Then the **direct product** of , , is the group whose underlying set is the Cartesian product of the underlying sets of , with multiplication defined componentwise. Thus, every element of is an -tuple , with .

Like the product topology, this has a universal property that characterizes it: it has a canonical projection homomorphism down to each , and any other group that projects to each with maps **factors through** the direct product: we can write for some . This is pretty easy to prove: let be the map that sends to the tuple whose th coordinate is . Then by definition.

Since groups have more structure than topological spaces, there are also injections sending to the tuple whose th coordinate is and all of whose other coordinates are identities. But the direct product *doesn’t* have the universal property for these injections. We’ll see a construction that does below. One way of noticing this is by seeing that the images of these injections don’t generate (since we defined generation in terms of finite products): they only generate the subgroup of whose elements have identities in all but finitely many of their coordinates.

Now, fortunately for our purposes, infinite products of groups are a lot less important than infinite products of topological spaces. More often, we’ll be working with finite products, and can rely on simple observations like with . As expected, is uniquely isomorphic to , etc., etc.

One thing of note is that contains both and as normal subgroups. (Check this.) In fact, there’s a way to go backwards: let be normal subgroups of with , . Then . Let’s prove this. The second condition allows us to write any element of as a product . Now suppose that ; then . Since , both products must be , so : any element of is a *unique* product of an element of and an element of . This shows that *as sets*.

But this isn’t enough, yet. Let’s look at the element , called the **commutator** of and , since it is equal to the identity if and only if and commute. We have , and similarly , so it is in , so it is the identity. Thus, every element of commutes with every element of . In particular, , which is precisely the multiplication of the direct product. Thus, .

As an exercise, prove that when and are relatively prime (they have no common factors other than .

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[…] seen how to find subgroups, and how to take the quotient by a normal subgroup, and how to find the direct product of a family of groups. Dual to the direct product is the free product, which generalizes the idea […]

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