Quotients and Homomorphisms for Beginners: Part 3, Nomenclature and the Integers Modulo n

The last post culminated in the definition of the quotient group. We saw that this group is constructed by essentially “dividing” the group into cosets and working with those. This is one of the reasons we deem the group the quotient group. One should note, however, that some authors use the alternative terminology factor group, e.g. Lang’s famous Algebra uses the term, but it is worth noting that he often uses odd conventions, for instance he introduces but never names the “Isomorphism theorems.”

There is another reason for the name “quotient,” which is probably where the term comes from in the first place. Before explaining the name, however, we should introduce the quintessential example of a quotient group: the integers modulo n. When working in general algebra, the integers are almost always the example we care about, because algebra was developed in no small part to do number theory. Let us quickly prove a theorem about the group of the integers under addition before proceeding.

Theorem. The subgroups of (\mathbb{Z}, +) are of the form n\mathbb{Z} for n an integer.

Proof. We will only prove that the subsets of that form are indeed subgroups. The proof that these are the only such subgroups will be left as an exercise (Hint: recall the Euclidean algorithm).

First, we show closure under addition. If nx, ny \in n\mathbb{Z}, then nx+ny=n(x+y), which is in n\mathbb{Z}. Clearly, the identity is 0= (0)n \in n\mathbb{Z}, and the inverse of n x is -nx=n(-x) \in n\mathbb{Z}. \blacksquare

It is easy to see that the subgroups n\mathbf{Z} are normal, because the subgroups are more than that: they are abelian. Therefore, we have the quotient group \mathbf{Z} / n\mathbf{Z} also known as the integers mod n and sometimes denoted \mathbf{Z}_n (not to be confused with the n-adic numbers). In my opinion, it is actually a little weird to see how this group fits in with quotient groups in general. Understanding the group is easy, but the notation and how we can figure out what it is just based on the general definition of G/H was not instant for me when I was first introduced to the subject. So, I will take a bit of care to show how the two coincide, in case I am not the only person susceptible to this little struggle. Either way, it will take some effort to truly internalize all of this.

Generally, we say \mathbf{Z}/n\mathbf{Z} := \{0, \ldots n-1 \} with addition mod n, but this is a bit misleading. Let us start from the beginning. The subgroup is n \mathbf{Z}=\{ 0 , \pm n , \pm 2n , \ldots \}, and so the cosets are

0+n\mathbf{Z} , 1+n\mathbf{Z}, \ldots , (n-1)+n\mathbf{Z}

We claim that these are all the cosets, because if k \in \mathbf{Z}, then

k=nq+r \; , \; 0 \leq r < n \implies k+n\mathbf{Z}=nq+n\mathbf{Z}+r= n \mathbf{Z} + r

But, this is already in the list, because of the restriction on r. \blacksquare

So, for simplicity we generally identify coset m + n\mathbf{Z} with m.

And, so we can compute the Cayley table (multiplication table) for small values of n quite easily. Below is the Cayley table for Z/4Z.

Z Mod 4Z Cayley

We can now see where the integers come in the naming of this group. When one divides, for instance, 15 by 5, one obtaains 3 because one can regroup 15 objects into 3 groups (in the colloquial sense) of 5 objects. In the quotient group, we have the same thing but with additional structure, namely the group structure of the integers.

Next time, we will digress a bit before continuing with the theory of quotient groups. Eventually, we will make our way to the isomorphism theorems and more.

Advertisements

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