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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 gG. 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 gG and hH, it will always be the case that gH=Hg.

Example 10.2.

Let H be the subgroup of S3 consisting of elements (1) and (12). Since

(123)H={(123),(13)}andH(123)={(123),(23)},

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)}.

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

(1) (2). Since N is normal in G, gN=Ng for all gG. Hence, for a given gG and nN, there exists an n in N such that gn=ng. Therefore, gng1=nN or gNg1N.

(2) (3). Let gG. Since gNg1N, we need only show NgNg1. For nN, g1ng=g1n(g1)1N. Hence, g1ng=n for some nN. Therefore, n=gng1 is in gNg1.

(3) (1). Suppose that gNg1=N for all gG. Then for any nN there exists an nN such that gng1=n. Consequently, gn=ng or gNNg. Similarly, NggN.

Subsection Factor Groups

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.

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 g1N is the inverse of gN. The order of G/N is, of course, the number of cosets of N in G.

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 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.

N(12)NNN(12)N(12)N(12)NN

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.

Example 10.6.

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,}.

The group Z/3Z is given by the Cayley table below.

+0+3Z1+3Z2+3Z0+3Z0+3Z1+3Z2+3Z1+3Z1+3Z2+3Z0+3Z2+3Z2+3Z0+3Z1+3Z

In general, the subgroup nZ of Z is normal. The cosets of Z/nZ are

nZ1+nZ2+nZ(n1)+nZ.

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.

Example 10.7.

Consider the dihedral group Dn, generated by the two elements r and s, satisfying the relations

rn=ids2=idsrs=r1.

The element r actually generates the cyclic subgroup of rotations, Rn, of Dn. Since srs1=srs=r1Rn, 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.