
# 5.0: Quotient Groups

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696
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Previously we saw product groups; now we'll learn about quotient groups. The construction is a bit more involved than the construction of product groups, just as division of natural numbers is a bit more complicated than multiplication...

Let's think for a moment about how quotients of natural numbers work, for the sake of building an imperfect analogy. When we write $$\frac{n}{d}=q$$, we have a numerator, denominator, and a quotient. The quotient $$q$$ can be thought of as the number of times we can divide $$n$$ into groups of $$d$$ objects.

In making a quotient group, then, we would like to start with a group $$G$$, identify a subgroup $$H$$ (the divisor) and do something to get a group $$G/\mathord H=Q$$. Using our analogy of dividing natural numbers, we would like to divide the group $$G$$ into collections according to $$H$$. The notion of coset does this quite nicely, and in fact previously allowed us to see that the order of any subgroup $$H$$ divides the order of $$G$$.

Definition 5.0.0

The set of cosets of a subgroup $$H$$ of $$G$$ is denoted $$G/\mathord H$$.

Then we can try to take the cosets of $$H$$ as the underlying set of our would-be quotient group $$Q$$. The question is whether we can now identify a reasonable group operation on the set of cosets of $$H$$. The answer is 'sometimes!'

#### A Bad Choice of Product on Cosets

Suppose we have two cosets of $$H$$, $$aH$$ and $$bH$$. We would like to define an operation on the two, so we naively write $$aH\cdot bH = abH$$, using the group operation in $$G$$ to multiply $$a$$ and $$b$$. And indeed, sometimes this works, but often it doesn't. What might cause it to fail? A problem arises because the set on which we're defining our new quotient group is the set of cosets, and it isn't generally obvious which element to take as the representative of the coset; ie, there is more than one way to write a coset as $$gH$$, and different choices might lead to different answers when we multiply our cosets.

Here's an example.

Take the group $$G=D_5$$, the symmetries of a pentagon generated by a flip $$f$$ and a rotation $$r$$. Let $$H$$ be the subgroup consisting of just the identity and the flip $$f$$. Then $$H=\{1, f\}$$. This subgroup has five different cosets; suppose we want to multiply the cosets $$C=\{ r, rf \}$$ and $$D=\{ r^3, r^3f \}$$. Notice that there are two different ways to write $$C$$ in the from $$gH$$: $$C=rH$$ and $$C=rfH$$. Each arises from a different choice of representative from $$H$$. The same is true for $$D$$: $$D=r^3H$$ and $$D=r^3fH$$. Depending on the choice of representatives, our rule for multiplying cosets then yields different answers. For example, $$rH \cdot r^3H = H$$, but $$rfH\cdot r^3H= rfr^3H = r^2fH\neq H$$.

Then we see that a more nuanced approach is necessary: in particular, our notion of a product shouldn't depend on a choice of coset representative!

Exercise 5.0.1

Let $$R$$ be the subgroup of $$D_5$$ generated by the rotation $$r$$. Show that the product rule $$(aR)\cdot (bR)$$ does not depend on the choice of coset representatives!

#### Products of Cosets

The initial idea for a product on cosets fell down because we were multiplying coset representatives, instead of thinking about how to multiply the actual cosets. So let's try to define an actual product of cosets!

Earlier, we saw what we might call left cosets, of the form $$aH = \{ah_1, ah_2, \ldots\}$$ where $$h_i$$ are all the elements of $$H$$. But we can easily imagine right cosets as well, $$Ha = \{h_1a, h_2a, \ldots\}$$, and even double cosets $$aHb=\{ah_1b, ah_2b, \ldots\}$$. More generally, we can define product of sets: if $$A=\{a_1, a_2, \ldots\}$$, then $$AH$$ is the set obtained by taking products of elements of $$A$$ and $$H$$ in every possible way: $$AH=\{ ah | a\in A, h\in H \}$$. We can use the product of sets to compute explicit products of cosets.

Proposition 5.0.2

If $$H$$ is a subgroup of $$G$$, then $$HH=H$$.

Proof 5.0.3

Since $$H$$ is closed under the group operation, every element of $$HH$$ is in $$H$$. Furthermore, since $$1\in H$$, every element $$h$$ in $$H$$ appears in $$HH$$ (for example, as $$1h$$). Then $$HH=H$$.

We build up these definitions so we can talk about products of cosets: $$(aH)(bH)$$. One fear of this approach is that taking a set-product like this may not give back a real left coset of $$H$$. In fact, sometimes it does and sometimes it doesn't!

If $$G$$ is a commutative group, then right cosets and left cosets are the same thing: $$ah_i=h_ia$$ for every $$h_i$$, so $$aH=Ha$$. In this case, when examining products like $$(aH)(bH)$$, we have $$aHbH=abHH=abH$$. Then defining a product on cosets $$(aH)(bH)=abH$$ makes sense, and will end up giving a nice group structure. The identity is $$1H$$, associativity follows from the multiplication rule in $$G$$, and inverses are easy: $$gHg^{-1}H=H$$.

We should also check that this product doesn't depend on choice of coset representative. Suppose $$aH=xH$$, and consider the product $$(aH)(bH)=aHHb=aHb$$. Then notice that $$(aH)(bH)=aHb=abH$$, and $$(xH)(bH)=xHb=(aH)b=abH$$. Thus, we have $$aH=bH$$.

#### Normal subgroups

If we look closely at what we've just done, we didn't actually need $$G$$ to be commutative: all that we needed was $$aH=Ha$$ for every $$a\in G$$. For example, we know this is true for the kernel of any homomorphism from the proposition in Section 4.2.

Definition 5.0.4: Normal Subgroups

A subgroup $$H$$ of a group $$G$$ is called a normal subgroup if $$aH=Ha$$ for every $$a\in G$$.

Then we've already proven the following theorem:

Theorem 5.0.5

Let $$H$$ be a normal subgroup of $$G$$. Then $$G\mathord H$$ is a group.

We'll also make explicit an earlier observation.

Proposition 5.0.6

Let $$G$$ be a commutative group. Then every subgroup $$H$$ of $$G$$ is a normal subgroup, and $$G/\mathord H$$ is a group.

We have already noticed that the kernel of any homomorphism is a normal subgroup. We can also define the quotient map $$\pi: G\rightarrow G/\mathord H$$, defined by $$\pi(a) = aH$$ for any $$a\in G$$. So long as the quotient is actually a group (ie, $$H$$ is a normal subgroup of $$G$$), then $$\pi$$ is a homomorphism. In fact, the kernel of $$\pi$$ is exactly $$H$$. So we observe:

Corollary 5.0.7

A subgroup of $$G$$ is normal if and only if it is the kernel of a homomorphism.

Exercise 5.0.8

Let $$G$$ be a finite group and $$H$$ a subgroup with $$\frac{|G|}{|H|}=2$$. Show that $$H$$ is a normal subgroup of $$G$$.

### Contributors

• Tom Denton (Fields Institute/York University in Toronto)