## Section 3.2 Factor Groups and Normal Subgroups

If \(H\) is a subgroup of a group \(G\text{,}\) then right cosets are not always the same as left cosets; that is, it is not always the case that \(gH = Hg\) for all \(g \in G\text{.}\) The subgroups for which this property holds play a critical role in group theory— they allow for the construction of a new class of groups, called factor or quotient groups. Factor groups may be studied directly or by using homomorphisms, a generalization of isomorphisms. We will study isomorphisms and homomorphisms in Chapter 4.

### Subsection 3.2.1 Normal Subgroups

A subgroup \(H\) of a group \(G\) is normal in G if \(gH = Hg\) for all \(g \in G\text{.}\) That is, a normal subgroup of a group \(G\) is one in which the right and left cosets are precisely the same.

###### Example 3.9.

Let \(G\) be an abelian group. Every subgroup \(H\) of \(G\) is a normal subgroup. Since \(gh = hg\) for all \(g \in G\) and \(h \in H\text{,}\) it will always be the case that \(gH = Hg\text{.}\)

###### Example 3.10.

Let \(H\) be the subgroup of \(S_3\) consisting of elements \((1)\) and \((12)\text{.}\) Since

\(H\) cannot be a normal subgroup of \(S_3\text{.}\) However, the subgroup \(N\text{,}\) consisting of the permutations \((1)\text{,}\) \((123)\text{,}\) and \((132)\text{,}\) is normal since the cosets of \(N\) are

The following theorem is fundamental to our understanding of normal subgroups.

###### Theorem 3.11.

Let \(G\) be a group and \(N\) be a subgroup of \(G\text{.}\) Then the following statements are equivalent.

The subgroup \(N\) is normal in \(G\text{.}\)

For all \(g \in G\text{,}\) \(gNg^{-1} \subset N\text{.}\)

For all \(g \in G\text{,}\) \(gNg^{-1} = N\text{.}\)

###### Proof.

(1) \(\Rightarrow\) (2). Since \(N\) is normal in \(G\text{,}\) \(gN = Ng\) for all \(g \in G\text{.}\) Hence, for a given \(g \in G\) and \(n \in N\text{,}\) there exists an \(n'\) in \(N\) such that \(g n = n' g\text{.}\) Therefore, \(gng^{-1} = n' \in N\) or \(gNg^{-1} \subset N\text{.}\)

(2) \(\Rightarrow\) (3). Let \(g \in G\text{.}\) Since \(gNg^{-1} \subset N\text{,}\) we need only show \(N \subset gNg^{-1}\text{.}\) For \(n \in N\text{,}\) \(g^{-1}ng=g^{-1}n(g^{-1})^{-1} \in N\text{.}\) Hence, \(g^{-1}ng = n'\) for some \(n' \in N\text{.}\) Therefore, \(n = g n' g^{-1}\) is in \(g N g^{-1}\text{.}\)

(3) \(\Rightarrow\) (1). Suppose that \(gNg^{-1} = N\) for all \(g \in G\text{.}\) Then for any \(n \in N\) there exists an \(n' \in N\) such that \(gng^{-1} = n'\text{.}\) Consequently, \(gn = n' g\) or \(gN \subset Ng\text{.}\) Similarly, \(Ng \subset gN\text{.}\)

### Subsection 3.2.2 Factor Groups

If \(N\) is a normal subgroup of a group \(G\text{,}\) then the cosets of \(N\) in \(G\) form a group \(G/N\) under the operation \((aN) (bN) = abN\text{.}\) This group is called the factor or quotient group of \(G\) and \(N\text{.}\) Our first task is to prove that \(G/N\) is indeed a group.

###### Theorem 3.12.

Let \(N\) be a normal subgroup of a group \(G\text{.}\) The cosets of \(N\) in \(G\) form a group \(G/N\) of order \([G:N]\text{.}\)

###### Proof.

The group operation on \(G/N\) is \((a N ) (b N)= a b N\text{.}\) This operation must be shown to be well-defined; that is, group multiplication must be independent of the choice of coset representative. Let \(aN = bN\) and \(cN = dN\text{.}\) We must show that

Then \(a = b n_1\) and \(c = d n_2\) for some \(n_1\) and \(n_2\) in \(N\text{.}\) Hence,

The remainder of the theorem is easy: \(eN = N\) is the identity and \(g^{-1} N\) is the inverse of \(gN\text{.}\) The order of \(G/N\) is, of course, the number of cosets of \(N\) in \(G\text{.}\)

It is very important to remember that the elements in a factor group are *sets of elements* in the original group.

###### Example 3.13.

Consider the normal subgroup of \(S_3\text{,}\) \(N = \{ (1), (123), (132) \}\text{.}\) The cosets of \(N\) in \(S_3\) are \(N\) and \((12) N\text{.}\) The factor group \(S_3 / N\) has the following multiplication table.

This group has order 2, and you might think it looks a lot like \({\mathbb Z}_2\) (in Chapter 4 we will see that this group is isomorphic to \(\Z_2\)). At first, multiplying cosets seems both complicated and strange; however, notice that \(S_3 / N\) is a smaller group. The factor group displays a certain amount of information about \(S_3\text{.}\) Actually, \(N = A_3\text{,}\) the group of even permutations, and \((12) N = \{ (12), (13), (23) \}\) is the set of odd permutations. The information captured in \(G/N\) is parity; that is, multiplying two even or two odd permutations results in an even permutation, whereas multiplying an odd permutation by an even permutation yields an odd permutation.

###### Example 3.14.

Consider the normal subgroup \(3 {\mathbb Z}\) of \({\mathbb Z}\text{.}\) The cosets of \(3 {\mathbb Z}\) in \({\mathbb Z}\) are

The group \({\mathbb Z}/ 3 {\mathbb Z}\) is given by the Cayley table below.

In general, the subgroup \(n {\mathbb Z}\) of \({\mathbb Z}\) is normal. The cosets of \({\mathbb Z } / n {\mathbb Z}\) are

The sum of the cosets \(k + n{\mathbb Z}\) and \(l + n{\mathbb Z}\) is \(k+l + n{\mathbb Z}\text{.}\) Notice that we have written our cosets additively, because the group operation is integer addition.

###### Example 3.15.

Consider the dihedral group \(D_n\text{,}\) generated by the two elements \(r\) and \(s\text{,}\) satisfying the relations

The element \(r\) actually generates the cyclic subgroup of rotations, \(R_n\text{,}\) of \(D_n\text{.}\) Since \(srs^{-1} = srs = r^{-1} \in R_n\text{,}\) the group of rotations is a normal subgroup of \(D_n\text{;}\) therefore, \(D_n / R_n\) is a group. The group has exactly two elements, so it will look just like \({\mathbb Z}_2\text{.}\)

### Reading Questions 3.2.3 Reading Questions

###### 1.

Give two ways we can can be sure that a subgroup \(H\) of a group \(G\) is normal.

###### 2.

What does the notation \(G/H\) represent? What does the notation \([G:H]\) represent? How are these related?

###### 3.

The subgroup \(8\Z\) is normal in \(\Z\text{.}\) In the factor group \(\Z/8\Z\text{,}\) perform the computation \((3+8\Z) + (7+8\Z)\text{.}\)

###### 4.

After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).

### Exercises 3.2.4 Practice Problems

###### 1.

Let \(G = D_4\text{,}\) the group of symmetries of a square, and take \(H = \{r_0, r_2, d_1, d_2\}\text{.}\) Is \(H\) normal in \(G\text{?}\) If so, how many elements are in the factor group \(G/H\text{?}\) Write out its group table.

###### 2.

A conjugate of an element \(x \in G\) is an element of \(G\) with the form \(gxg\inv\) for some \(g \in G\text{.}\) Using \(G = D_4\) and \(H = \{r_0, r_2, d_1, d_2\}\text{,}\) find all the conjugates of all the elements of \(H\text{.}\) What do you notice?

###### 3.

Prove that if \(G\) is abelian and \(H\) is a subgroup, that \(G/H\) is abelian.

###### 4.

True or false: If \(H\) is abelian and \(G/H\) is abelian, then \(G\) is abelian. Prove your answer.

Look at Exercise 3.2.4.1 .

###### 5.

Let \(G = D_4\) and \(H = \{r_0, h\}\text{.}\) Prove that \(H\) is not normal in \(G\text{.}\) Then give a specific example of what goes wrong if you try to define coset multiplication on the set of cosets \(G/H\text{.}\)

###### 6.

Again let \(G = D_4\) and \(H = \{r_0, h\}\text{.}\) Show that \(H \) is not closed under conjugates.

### Exercises 3.2.5 Collected Homework

###### C1.

Let \(G = S_3\) and \(H = \{(1), (13)\}\text{.}\)

Find the three (left) cosets of \(H\) in \(G\text{.}\)

Explain what goes wrong if you try to define an operation on the set of cosets \(G/H\text{.}\) Do this with a specific example using this \(G\) and \(H\text{.}\)

Write out all the conjugates of \(H\) in \(G\text{.}\) That is, find \(gxg\inv\) for all \(x \in H\) and \(g \in G\text{.}\) Is \(H\) closed under conjugates?

Show that \(H\) is not normal in \(G\) using the definition (about left and right cosets).

###### C2.

BONUS! Prove that if \([G : H] = 2\) then \(H\) is normal in \(G\text{.}\)

This is not all that hard. Try an example (such as Exercise 3.2.4.1 ) to see why this works. Then explain it in general.