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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 10.1 Factor Groups and Normal Subgroups

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Subsection Normal Subgroups

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A subgroup H of a group G is normal in G if gH=Hg for all g∈G. That is, a normal subgroup of a group G is one in which the right and left cosets are precisely the same.
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Example 10.1.

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Let G be an abelian group. Every subgroup H of G is a normal subgroup. Since gh=hg for all g∈G and h∈H, it will always be the case that gH=Hg.
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Example 10.2.

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Let H be the subgroup of S3 consisting of elements (1) and (12). Since
(123)H={(123),(13)}andH(123)={(123),(23)},
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H cannot be a normal subgroup of S3. However, the subgroup N, consisting of the permutations (1), (123), and (132), is normal since the cosets of N are
N={(1),(123),(132)}(12)N=N(12)={(12),(13),(23)}.
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The following theorem is fundamental to our understanding of normal subgroups.

Proof.

(1) ⇒ (2). Since N is normal in G, gN=Ng for all g∈G. Hence, for a given g∈G and n∈N, there exists an n′ in N such that gn=n′g. Therefore, gng−1=n′∈N or gNg−1⊂N.
(2) ⇒ (3). Let g∈G. Since gNg−1⊂N, we need only show N⊂gNg−1. For n∈N, g−1ng=g−1n(g−1)−1∈N. Hence, g−1ng=n′ for some n′∈N. Therefore, n=gn′g−1 is in gNg−1.
(3) ⇒ (1). Suppose that gNg−1=N for all g∈G. Then for any n∈N there exists an n′∈N such that gng−1=n′. Consequently, gn=n′g or gN⊂Ng. Similarly, Ng⊂gN.
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Subsection Factor Groups

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

Proof.

The group operation on G/N is (aN)(bN)=abN. 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. We must show that
(aN)(cN)=acN=bdN=(bN)(dN).
Then a=bn1 and c=dn2 for some n1 and n2 in N. Hence,
acN=bn1dn2N=bn1dN=bn1Nd=bNd=bdN.
The remainder of the theorem is easy: eN=N is the identity and g−1N is the inverse of gN. The order of G/N is, of course, the number of cosets of N in G.
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It is very important to remember that the elements in a factor group are sets of elements in the original group.
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Example 10.5.

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Consider the normal subgroup of S3, N={(1),(123),(132)}. The cosets of N in S3 are N and (12)N. The factor group S3/N has the following multiplication table.
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N(12)NNN(12)N(12)N(12)NN
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This group is isomorphic to Z2. At first, multiplying cosets seems both complicated and strange; however, notice that S3/N is a smaller group. The factor group displays a certain amount of information about S3. Actually, N=A3, 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.
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Example 10.6.

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Consider the normal subgroup 3Z of Z. The cosets of 3Z in Z are
0+3Z={…,−3,0,3,6,…}1+3Z={…,−2,1,4,7,…}2+3Z={…,−1,2,5,8,…}.
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The group Z/3Z is given by the Cayley table below.
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+0+3Z1+3Z2+3Z0+3Z0+3Z1+3Z2+3Z1+3Z1+3Z2+3Z0+3Z2+3Z2+3Z0+3Z1+3Z
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In general, the subgroup nZ of Z is normal. The cosets of Z/nZ are
nZ1+nZ2+nZ⋮(n−1)+nZ.
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The sum of the cosets k+nZ and l+nZ is k+l+nZ. Notice that we have written our cosets additively, because the group operation is integer addition.
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Example 10.7.

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Consider the dihedral group Dn, generated by the two elements r and s, satisfying the relations
rn=ids2=idsrs=r−1.
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The element r actually generates the cyclic subgroup of rotations, Rn, of Dn. Since srs−1=srs=r−1∈Rn, the group of rotations is a normal subgroup of Dn; therefore, Dn/Rn is a group. Since there are exactly two elements in this group, it must be isomorphic to Z2.
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