Lecture 6: Quotient Groups

6 Normal Subgroups and Quotient Groups

6.1 Review
In the last lecture, we learned about the correspondence theorem.

Theorem 6.1 (Correspondence Theorem)

Where f : G → G′ is a surjectivea homomorphism, and K = ker(f ), there is a correspondence

{subgroups H : K ⊆ H ⊆ G} ↭ {subgroups H ′ : {eG′ } ⊆ H ′ ⊆ G′ }},

which states that subgroups of G containing the kernel are in bijection with subgroups of H ′ in the image
of f.
a In fact, it is possible to slightly revise the statement of the theorem such that the surjective condition is no longer necessary.

The correspondence comes from taking

H 7→ f (H) ⊆ G′
{x ∈ G : f (x) ∈ H ′ } = f −1 (H ′ ) ←[ H ′

From the point of view of understanding subgroups, the correspondence theorem allows us to understand a
slice of G. If G is a complicated group with many surjective maps onto diferent groups G′ , we can use the
correspondence theorem multiple times to understand G, and conversely, if G′ is a complicated group, we can
use G to study G′ .

Proof. In order to show that this correspondence is a bijection, we check that these are inverses to each other.
• If K ⊆ H ⊆ G, then we want to show that f −1 (f (H)) = H. By defnition,
f −1 (f (H)) = {x ∈ G : f (x) = f (h), h ∈ H}.
By defnition, H ⊆ f −1 (f (H)). Also, if x ∈ f −1 (f (H)), then f (x) = f (h) for some h ∈ H. This is true if
and only if x is in the coset hK; in other words, x = h · k for some k ∈ K. Since K ⊆ H,24 k ∈ H, and so
x = hk ∈ H.
• The proof of the other direction is left as an exercise to the reader (it is very much the same idea).

6.2 Normal Subgroups

Recall the defnition of a normal subgroup.

Defnition 6.2
A subgroup H ⊆ G is normal if xHx−1 = H for all x ∈ G.

The notation H ≤ G denotes that H is a subgroup, not just a subset, of G. Now, the notation H ⊴ G will
denote that H is a normal subgroup of G.25

Example 6.3 (Kernel)

The kernel ker(f ) is always normal.

Guiding Question
Given any normal subgroup N ⊴ G, is there always a group homomorphism f : G → G′ such that
N = ker(f )?
Answer: Yes!

24 The theorem is not true if the kernel is not contained in H, so this fact must be used at some point.
25 This notation will not necessarily be used consistently throughout this lecture/class, but it is used in the literature.

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Lecture 6: Quotient Groups

Let’s look at a quick example frst.

Example 6.4 (Integers modulo 2)

If G = Z and H = 2Z, the homomorphism is
f
→ G′ = Z2
G−
n 7→ n mod 2.

The kernel of f consists of the elements mapping to 0 mod 2; that is, even integers, which is precisely 2Z.

In the case when N = ker(f ), the cosets26 of N are in correspondence with im(f ), by the correspondence
theorem. Since they are in bijective correspondence, the group structure on im(f ) can be carried over to the
set of cosets of N.

6.3 Quotient Groups

Now that we’ve defned cosets, we have the following question:

Guiding Question
Can we directly defne a group structure on the sets of cosets of N ?

If C1 , C2 ⊆ G are cosets, what should C1 · C2 be? The most intuitive defnition would be to take the set of
products of each of the elements:

Defnition 6.5
Let the product structure on the cosets be defned as

C1 · C2 := {x ∈ G : x = y1 · y2 ; y1 ∈ C1 , y2 ∈ C2 },

the pairwise product.

Theorem 6.6
If C1 , C2 are cosets of a normal subgroup N , C1 · C2 is also a coset of N.

It is crucial that N is normal!

Example 6.7
Consider H = {e, y} ⊆ G = S3 . Then H is not a normal subgroup. Consider xH = {x, xy}. We have

xH · xH = {x2 , x2 y, xyx = y, xyxy = e},

which is not a coset!

Proof. Let C1 = aN and C2 = bN.

• The inclusion abN ⊆ C1 · C2 holds because abn = (ae)(bn) ∈ C1 · C2 , since ae ∈ C1 and bn ∈ C2 .
26 The left and right cosets are the same when N is normal.

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Lecture 6: Quotient Groups

• Take an1 · bn2 ∈ C1 · C2 . Since N is normal, bN = N b, so n1 · b = b · n3 for some n3 ∈ N. So

an1 · bn2 = abn3 n2 ∈ abN.

Then C1 · C2 = abN.

So it is only when N is normal that we do in fact have a product structure on the set of cosets of N !

Defnition 6.8
The quotient group G/N is the set of cosets of a normal subgroup N. The group structure is defned asa

[C1 ] · [C2 ] := [C1 · C2 ]

[aN ] · [bN ] := [abN ].

The right hand side is a coset because N is a normal subgroup.b

a The notation [x] refers to the equivalence class of x under an equivalence relation; in this case, the equivalence relation is

defned by the partition of G into cosets.

b The product can be verifed to be independent of the representatives a and b from the fact that N is normal.

Theorem 6.9
The following two statements are true about the quotient group:
1. The composition law, as defned in Defnition 6.8 does defne a group structure on G/N (all the group
axioms hold).
2. There exists a surjective homomorphism

π : G → G/N
x 7→ [xN ]

such that ker(π) = N.

This is one of the most basic operations we can do on groups!

Proof. First of all, let’s show that G/N is actually a group.

• Identity. The identity is [N ] = [eN ]. The product is

[aN ] · [N ] = [aeN ] = [aN ].

• Inverse. We can check that

[aN ]−1 = [a−1 N ].

In general, the inverse of a left coset will be a right coset, but because N is normal, they are the same.
• Associativity. Similarly, associativity of G/N boils down to associativity for G.27
Now, we can show the second part of the theorem. Take π(x) = [xN ]. It is evidently a surjective map. Then

π(xy) = [xyN ] = [xN ] · [yN ] = π(x) · π(y).

Then the kernel is

ker(π) = {x ∈ G : x ∈ N } = N.

27 The proof is left as an exercise for the reader.

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Lecture 6: Quotient Groups

Most of the proof of Theorem 6.9 seems very tautological. In fact, most of the action happened earlier on, in
Thereom 6.5, which showed that the product of two cosets actually was another coset, demonstrating that the
group structure does makes sense.
Here is an example of how this theorem is often used.

Example 6.10 (Quotient Group of SL2 (R))

Take N = {±I2 } ⊴ G = SL2 (R). Then, taking the quotient group SL2 (R)/{±I2 } gives a new group
P SL2 (R). Thus, from an explicitly defned group, in this case SL2 (R), we obtain a new, potentially
interesting or useful group by taking a quotient.

Another perspective on G/N is that it is similar to modular arithmetic. We have that a ≡ b mod N if
aN = bN ⊆ G. 28

6.4 First Isomorphism Theorem

f
Suppose we start of with G − → G′ a surjective homomorphism, and assume K = ker(f ) is a normal subgroup.
Given that it is a normal subgroup, we can feed it into this machine that we have created. Then

π : G → G/K

is a surjective group homomorphism. So we have started with a surjective group homomorphism and created
another surjective group homomorphism. But in fact, we have done nothing at all! There exists an isomorphism
∼
f¯ : G/K −
→ G′ .

The diagram
f
G G′
π
f
G/K
commutes.
So f = f¯ ◦ π. So up to isomorphism, our original group homomorphism is the same as our new one. This is not
surprising, because there is a correspondence between cosets of the kernel and points in the image. All we are
saying is that that bijection is compatible with the group structures on both sides. So f¯([xk]) = f (x). This is
known as the First Isomorphism Theorem.

28 We placed an equivalence relation on the group, and placed a group structure on the equivalence classes.

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MIT OpenCourseWare
https://ocw.mit.edu

Resource: Algebra I Student Notes

Fall 2021
Instructor: Davesh Maulik
Notes taken by Jakin Ng, Sanjana Das, and Ethan Yang

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