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With Respect to a Normal Subgroup |
The conceptual definitions of mathematical groups and subgroups involve no intellectual surprises. The notion of a factor or quotient group is a different matter. The factor group of a group G with respect to a special subgroup N, called a normal subgroup, can best be thought of a partition of the group G into classes over which a group operation can be unambiguously defined.
To explain these concepts it is best to start with the concept of cosets which are defined for any subgroup. Let H be a subgroup of G. The left coset of H for any element x of G is the set xH = {xh: h∈H }. Likewise the right coset of H for any element y of G is the set Hy = {hy: h∈H }. At this point it is convenient to deal with just the left cosets but the same things apply for right cosets.
One interesting thing about cosets is that they partition the set G into mutually exclusive subsets.
Proof: Consider two left cosets xH and yH. Suppose xH and yH have a common element z. This means that there exists h_{1} and h_{2} belonging to H such that z=xh_{1} and z=yh_{2}. Thus xh_{1}=yh_{2} and hence y = xh_{1}h_{2}^{-1}. Now let w be any element of yH. The there exists h_{3} such that w=yh_{3}. But this implies that w=(xh_{1}h_{2}^{-1})h_{3}=x(h_{1}h_{2}^{-1}h_{3}). Since h_{1}h_{2}^{-1}h_{3} necessarily belongs to H w must belong to xH. Thus yH ⊆ xH. Likewise a similar argument demonstrates that xH ⊆ yH so xH=yH.
There are two basic results for use later.
Proof: Suppose that both a and a' belong to xH. Then there exist h_{1} and h_{2} belonging to H such that a=xh_{1} and a'=xh_{2}. Since a=xh_{1}, x=ah_{1}^{-1} so a'=(ah_{1}^{-1})h_{2}=a(h_{1}^{-1}h_{2}). Thus a'=ah where h=h_{1}^{-1}h_{2}.
Proof: Since a'=ah this means a'∈aH. Since a∈aH then a and a' are in the same left coset; i.e., aH.
The same results applies to right cosets.
Normal subgroups are subgroups for which the partition of G is the same for right cosets as for left cosets. In other words, each left coset is also a right coset and vice versa.
Proof: Let a be any element of G. Then the left coset aN is also a right coset which means there exists an element of b such that aN=Nb. Since a belongs to aN it also belongs to Nb. But a also belongs to Na since the identity is an element of N. Therefore a is a common element of Na and Nb. By Lemma 0 this means Na=Nb. But is Na=Nb then Na=aN. So for any element a of G, aN=Na.
Proof: aN=Na means that for any n_{1} there exists n_{2} such that an_{1}=n_{2}a. Thus an_{1}a^{-1}=n_{2}; i.e., an_{1}a^{-1}=n_{2} ∈ N. This means that aN_{1}a^{-1} ⊆ Likewise for any n choose n_{3} to be a^{-1}na so an_{3}a^{-1}= a(a^{-1}na)a^{-1} is n and hence n belongs to aNa^{-1} and thus N ⊆ aNa^{-1}. This means aNa^{-1}=N.
For a normal subgroup a group operation can be defined on the cosets. For any two cosets A and B the group product of A with B is defined as the coset of the group product of ab where a is any element of A and b is any element of B. The crucial matter here is proving that no matter which representatives of A and B are chosen the coset of the product is the same.
Let A and B be any two cosets of N and a and b elements of A and B, respectively. Therefore A=Na and B=bN. Let C be the coset of c=ab. Now consider a' and b' be any other elements of aN and bN besides a and b. It must be shown that the coset of a'b' is the same as the coset of ab. From Lemma 1 there must exist h_{a} and _{b} such that a'=h_{a}a and b'=bh_{b}. Therefore a'b' = h_{a}abh_{b} and consequently a'b'h_{b}^{-1}=h_{a}ab. This means that a'b'h_{b}^{-1} belongs to the same coset as ab. But a'b'h_{b}^{-1} belongs to the same coset as a'b' so a'b' belongs to the same coset as ab. Therefore choosing any other representive of the cosets A and B defines the same group product for the cosets.
It remains to be established that the product operation defined above is in fact a group operation. The group identity for the cosets is the normal subgroup N, which is the coset of the identity element e of N. Thus the identity element e can be chosen as the representive of the coset N. The inverse for the coset aN is the coset of a^{-1}. There is associativity of the group operation for the cosets because of the associativity of the group operation for G; i.e., (aNbN)cN = aN(bNcN) because (ab)c=a(bc).
Thus the following theorem has been established:
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