## Group Theory Primer, part 3: direct products, cosets, and quotients!

### June 20, 2010

If we think about groups as if they were numbers, we’d want to add, subtract, multiply, and divide stuff.  Unfortunately, groups aren’t as simple as numbers, and we have more complex notions of what all of these things should correspond to.

## Direct Product.

Corresponding to addition of numbers, we have the idea of the direct product of groups, and this is a relatively easy idea to define.  Given two groups, say $(A, \cdot)$ and $(B, +)$ where each is paired with the operation in the group, we have that the direct product of groups is given by $A\times B = \{(a,b)\, |\, a\in A, b\in B\}$.  In other words, we just take all the ordered pairs where the first coordinate is an element of $A$ and the second coordinate is an element of $B$.  But if this is going to be a group, we need some way to manipulate the elements!  So, let’s take two elements in $A\times B$; let’s call them $(a,b)$ and $(c,d)$.  Let’s define the operation $\times$ such that $(a,b)\times (c,d) = (a\cdot c, b + d)$.  We simply use the operations we’ve defined in each set to manipulate the elements coordinate-wise.  Neat.

If $A$ and $B$ are groups, is $A\times B$ a group with that relation we defined above?  Yeah, but it’s not obvious.  Or, maybe it is!  What’s the identity element?  $(id_{A}, id_{B})$ where $id_{A}$ is the identity element of the group $A$ and similarly for $id_{B}$.  What’s the inverse for $(a,b)$?  Why, it’s simply $(a^{-1}, b^{-1})$.  Think about it.  Scratch it into some paper.  Yes, this is actually a group!  Oh, wait, is it closed under the operation?  Yes, because each group is, and we’re manipulating each of the elements component-wise.

## Cosets.

I guess I should have defined subgroups at some point, but the idea is so easy that I waited until I absolutely needed it before I defined it!  A subgroup is simply a set of elements inside of a group that functions as its own little group.  Specifically, it contains the identity, it is closed under the operation of the original set, and it contains all of its inverses.  We write $H\subseteq G$ to say that $H$ is a subset of $G$.  Cute.  For example, the even integers (under addition) are a subgroup of the integers (under addition).  Every group has the trivial subgroup which consists of just the identity, and every group has the improper subgroup which consists of the entire group (this is kind of silly, but it is sometimes useful to think of a group as a subgroup of itself).  Every non-trivial group which is not the whole group is called a proper subgroup.

Now, let’s talk about cosets for a second.  Specifically, we’re going to talk about left cosets.  What happens if we consider a subgroup $H\subseteq G$ and we multiply the subgroup by some element of the group.  What the hell do I mean by this?!  Let’s do this with an example.

Suppose our entire group $G={\mathbb Q}$ and our subgroup is the even integers, $H = 2{\mathbb Z}$.  So, what happens if we multiply our subgroup by, say, $\frac{1}{2}$?  It would look like this: $\{\frac{1}{2}h\,|\, h\in H\}$.  In other words, we’d get the set of all elements that are in $H$ multiplied by $\frac{1}{2}$.  If you’re clever, you’ll notice that this is, in fact, the integers!  Now, what if we multiplied $H$ by something else?  Like 2?  Then we get $2H = \{2h\, |\, h\in H\}$; or, in other words, we get every element of $H$ multiplied by 2.  The clever reader will notice that this is exactly the negative and positive multiples of 4.  Coo’.

The left coset is, then, defined the same way.  Given $x\in G$, we have that, for a subgroup $H\subseteq G$, we have that $xH = \{xh\, |\, h\in H\}$.

Who cares, you might be screaming.  Who would care about such a structure!?  Well, it just so happens that we use this to define one of the most important notions in group theory: the quotient.

## Quotients, at last!

The quotient is a tricky monster.  In my first pass at algebra, it was the operation that tricked me up the most.  It took a lot of thinking about to really get an intuitive idea of what exactly quotienting in groups was, and how it was at all related to the quotients of numbers that we know and love.  It’s essentially the same, after all: for some real number, dividing by something means “splitting a cake into this many pieces”, but for groups it kind of means, “split a cake into pieces that all look similar to the one we’re dividing out by.”  This is probably meaningless to you now, since we’ve not defined it yet, but come back to it after you’ve thought about quotienting and it might make more sense.  Maybe not!

Now, there are a number of ways to define the quotient, and a lot of them have to do with this weird picture of lines jumping down to dots and things, but I think the more formal way is an okay way to think about quotients for now.

Let’s define $G/H = \{x\cdot H\, |\, x\in G\}$, where $\cdot$ is the operation in $G$.  That’s it.  That’s all there is.  We do need to define the operation on this set to make it a group, but we’ll just use the ordinary operation that $G$ is using.   Then, given $x\cdot H$ and $y\cdot H$, we define $(x\cdot H)\cdot (y\cdot H) = (x\cdot y)\cdot H$.  In other words, to operate on cosets, it’s enough to operate on the elements that define which coset we’re using.

(It does not suffice to just say this, though.  The angry or clever reader will notice that it is possible for two different elements to define the same coset (for example, the coset $h_{1} + H$ and $h_{2} + H$ will be the same if $h_{1}, h_{2}\in H$).  So why should it suffice to pick $h_{1}$ rather than $h_{2}$, and why don’t we get a different answer?  This question, while valid, will be addressed later — just take my word for it now that, if the operation is reasonably nice, nothing bad will happen if you pick and choose from any element defining the coset when combining cosets together.)

We say that this is the quotient group or we say this is the group G modulo H, which is a significantly cooler way to say it.  It’s common, aloud, to read this “G mod H.” and every math student will know what you mean by this, so long as she knows that $G$ and $H$ are groups.

So what is this nonsense?  $G/H$ is a group (we have not proved this!  but it is true if $H$ is normal,) whose elements consist of the left cosets of $H$ in $G$.  Remember cosets?  They’re kind of like “multiples of $H$ in the group $G$“, and, in this case, they make up the quotient group.  Let’s take an example so that we feel a little better about this.  Note that the next two examples use SPECIAL subgroups called normal subgroups that we’ll talk about in the next section, and that’s why the coset manipulations work out so nicely.  For now, don’t worry yourself too much about this; we’ll talk about it in the next section.

Let’s let $G = {\mathbb Z}$ under addition and let’s let $H = 4{\mathbb Z}$ under addition; or, in other words, $H$ is all the integer multiples of 4.  What do we get when we divide out by this junk?  By our definition, we have $G/H = {\mathbb Z}/4{\mathbb Z} = \{x + 4{\mathbb Z}\, |\, x\in {\mathbb Z}\}$.  Well, what do the elements look like?  Here are some of the elements in $G/H$:

$\{0 + 4{\mathbb Z}, 1 + 4{\mathbb Z}, 2 + 4{\mathbb Z}, 3 + 4{\mathbb Z}, 4 + 4{\mathbb Z}, 5 + 4{\mathbb Z}, \dots \}$.

What are these cosets?  Well,

$0 + 4{\mathbb Z} = \{\dots, -8, -4, 0, 4, 8, \dots\}$

$1 + 4{\mathbb Z} = \{\dots, -7, -3, 1, 5, 9, \dots\}$

$2 + 4{\mathbb Z} = \{\dots, -6, -2, 2, 6, 10, \dots\}$

$3 + 4{\mathbb Z} = \{\dots, -5, -1, 3, 7, 11, \dots\}$

$4 + 4{\mathbb Z} = \{\dots, -4, 0, 4, 8, 12, \dots\}$

$5 + 4{\mathbb Z} = \{\dots, -3, 1, 5, 9, 13, \dots\}$

and so on.  Now, notice something here…a lot of these are actually the same!  $0 + 4{\mathbb Z}$ is the same as $4 + 4{\mathbb Z}$, $1 + 4{\mathbb Z}$ is the same as $5 + 4{\mathbb Z}$, and so on!  We actually only have four different cosets, because all of the other ones overlap!  Therefore, the only distinct elements in our quotient group are:

$G/H = \{0 + 4{\mathbb Z}, 1 + 4{\mathbb Z}, 2 + 4{\mathbb Z}, 3 + 4{\mathbb Z}\}$

ALSO, notice here that if we want to add these cosets together, it suffices to add the number before them together.  For example, let’s check that $(2 + 4{\mathbb Z}) + (1 + 4{\mathbb Z})$ is something reasonably nice.  In fact, if we take all the elements from both of them, we see that the first will be in the form $2+4n$ for some integer $n$, and the second will be $1 + 4m$ for some integer $m$.  Adding these, we get $(2 + 4n) + (1 + 4m) = 3 + 4(n+m)$, which is, as you can check, an element of $3 + 4{\mathbb Z}$.  Interestingly enough, if we use the defined operation above (when we defined quotient groups) then we note that all we needed to do was add that little 2 and that little 1 in the beginning of each coset to give us exactly what we wanted: $(2 + 4{\mathbb Z}) + (1 + 4{\mathbb Z}) = (2 + 1) + 4{\mathbb Z} = 3 + 4{\mathbb Z}$.  Nice.

By the way, we call this structure ${\mathbb Z}_{4}$, or “the integers modulo 4” or, sometimes, just “Z modulo 4.”  We know what you mean when you say it!  The reason for this is, if we talk about the cosets as if they were just the numbers in front of the coset, $\{0,1,2,3\}$, and we add them using coset addition, we notice that things like $3 + 2 = 1$ happen, which is the same as if we were to have arithmetic modulo 4.  If you’re not sure what this means, don’t worry about it; we’ll go over it later.  Or: look up “clock arithmetic” or “modular arithmetic.”  Stop being so lazy and do it!

Why don’t we try another example?  Just for kicks?  Do you remember $D_{8}$?  Well, we have that it consists of a bunch of rotations and reflections.  As you recall, $D_{8} = \{1, r, r^2, r^3, sr, sr^2, sr^3\}$, with composition as the operation.

Let’s define a subgroup of $D_{8}$.  Let’s call it $H = \{1, r^2\}$ .  Now let’s compute $G/H$!  Okay, so, what is this actually equal to?  $G/H = \{xH\, |\, x\in D_{8}\}$

So what are the cosets?  Well, we have…

$1\{1, r^2\} = \{1, r^2\}$

$r\{1, r^2\} = \{r, r^3\}$

$r^2\{1, r^2\} = \{r^2, 1\}$

$r^3\{1, r^2\} = \{r^3, r\}$

$s\{1, r^2\} = \{s, sr^2\}$

$sr\{1, r^2\} = \{sr, sr^3\}$

$sr^{2}\{1, r^2\} = \{sr^{2}, s\}$

$sr^{3}\{1, r^2\} = \{sr^{3}, sr\}$

And so the only distinct cosets are the following:

$G/H = \{\{1, r^2\}, \{r, r^3\}, \{s, sr^2\}, \{sr, sr^3\}\}$

And that is exactly the set of cosets.  As you can see, we have nice little slices that all look like $\{1, r^2\}$, which is kind of cute.

Notice, also, that each element is represented exactly once in each of the cosets.  Cute.  This is a theorem, but we’ll assume that it’s true.  That means we can represent each of the cosets in the quotient group by just one element in their coset!  Okay, so, if we do this, we can say that:

$G/H = \{\overline{1}, \overline{r}, \overline{s}, \overline{sr}\}$

where we put the little bar above them to note that they’re representing their entire coset.

## Normal Subgroups.

(Edit: Thanks to Brk who pointed out that we cannot do nice quotients and things without using normal subgroups.)

Remember before when I said quotients were groups if we had that the subgroup was a special kind of subgroup called a normal subgroup?  Well, we might as well go over them now, lest you start thinking we can just willy-nilly divide by whatever you want and have it be suitably nice.

You know what’s the worst?  When you try to divide a whole number by something and it doesn’t evenly divide.  What happens when we divide 5 by 2?  We get some kind of crappy fraction, and that sucks!  We don’t want the same thing to happen in our groups.  But we have to be clever.

When we had $G/H$, we had a collection of cosets.  So, let’s call one of the cosets $x\cdot H$ for some $x\in G$.  Now, if we wanted to manipulate this to get back to $H$, the coset, we’ll need to multiply this by the inverse of $x$.  So, we’ll have $x\cdot H\cdot x^{-1}$.  Now, look at this…how could this fail to be equal to $H$?  In other words, how could we not have $x\cdot H\cdot x^{-1} = H$?  Well, there’s a number of good examples, but let’s just do one really easy one: suppose that $H$ was a subset that had just one invertible matrix in it.  Then it is almost never the case that $M\cdot H\cdot M^{-1} = H$ for any matrix $M$.  Upsetting!  Well, we’d like to say, when we divide things out, that $x\cdot H\cdot x^{-1} = H$ for any $x\in G$.

Well, this kind of subgroup is really nice and it has some nice properties, too!  We should probably call it normal.  In fact, we do just that.  Specifically, we call a subgroup $H$ normal if and only if we have $x\cdot H\cdot x^{-1} = H$.  There are many other characterizations of normal subgroups, but this is the nicest looking one!

When we have normal subgroups, that coset manipulation that we did above in the quotient group is actually defined and it actually becomes a group!  Normal subgroups are useful for many things, but there’s a lot of cool quotient stuff.

In the next post, we’ll go over what else makes a subgroup normal, and we’ll go over how to represent quotient groups nicely.  Similarly, we’ll go over a way to mathematically say, “wow, these groups look really similar!” and a few elementary theorems of group theory.