Gracious Living


Products of Groups
November 19, 2010, 22:49
Filed under: Algebra, Math | Tags: , , , ,

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 G_i be a family of groups indexed by i\in I.  Then the direct product of G_i, \prod_{i\in I}G_i, is the group whose underlying set is the Cartesian product of the underlying sets of G_i, with multiplication defined componentwise.  Thus, every element of \prod G_i is an I-tuple (g_i)_{i\in I}, with (g_i)(h_i)=(g_ih_i).

Like the product topology, this has a universal property that characterizes it: it has a canonical projection homomorphism p_i down to each G_i, and any other group H that projects to each G_i with maps q_i factors through the direct product: we can write q_i=p_i\circ r for some r:H\rightarrow\prod G_i.  This is pretty easy to prove: let r be the map that sends h\in H to the tuple whose ith coordinate is q_i(h).  Then p_ir(h)=q_i(h) by definition.

Since groups have more structure than topological spaces, there are also injections G_i\rightarrow\prod G_i sending g\in G_i to the tuple whose ith coordinate is G 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 \prod G_i (since we defined generation in terms of finite products): they only generate the subgroup of \prod G_i 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 G\times H=\{(g,h):g\in G,h\in H\} with (g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2).  As expected, (G\times H)\times J is uniquely isomorphic to G\times (H\times J), etc., etc.

One thing of note is that G\times H contains both G and H as normal subgroups.  (Check this.)  In fact, there’s a way to go backwards: let G,H be normal subgroups of J with G\cap H=\{e\}, GH=J.  Then J\cong G\times H.  Let’s prove this.  The second condition allows us to write any element of J as a product gh,g\in G,h\in H.  Now suppose that j=g_1h_1=g_2h_2; then g_2^{-1}g_1=h_2h_1^{-1}.  Since G\cap H=\{e\}, both products must be e, so g_1=g_2,h_1=h_2: any element of J is a unique product of an element of G and an element of H.  This shows that J\cong G\times H as sets.

But this isn’t enough, yet.  Let’s look at the element ghg^{-1}h^{-1}, called the commutator of g and h, since it is equal to the identity if and only if g and h commute.  We have ghg^{-1}h^{-1}\in G(hGh^{-1})=GG=G, and similarly ghg^{-1}h^{-1}\in (gHg^{-1})H=HH=H, so it is in G\cap H, so it is the identity.  Thus, every element of G commutes with every element of H.  In particular, g_1h_1g_2h_2=g_1g_2h_1h_2, which is precisely the multiplication of the direct product.  Thus, J\cong G\times H.

As an exercise, prove that \mathbb{Z}_m\times\mathbb{Z}_n\cong\mathbb{Z}_{mn} when m and n are relatively prime (they have no common factors other than 1.

Advertisements

1 Comment so far
Leave a comment

[…] 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 […]

Pingback by More Products of Groups « Gracious Living




Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s



%d bloggers like this: