## Section 10.1 Factor Groups and Normal Subgroups

### Subsection 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 10.1.

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 10.2.

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 10.3.

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 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 10.4.

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 10.5.

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 is isomorphic to \({\mathbb Z}_2\text{.}\) 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 10.6.

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 10.7.

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. Since there are exactly two elements in this group, it must be isomorphic to \({\mathbb Z}_2\text{.}\)