Definition
Let N be a normal subgroup of a group G. We define the set G/N to be the set of all left cosets of N in G, i.e., G/N = { aN : a in G }. The group operation on G/N is the product of subsets defined above. In other words, for each aN and bN in G/N, the product of aN and bN is (aN)(bN). This operation is closed, because (aN)(bN) really is a left coset:
- (aN)(bN) = a(Nb)N = a(bN)N = (ab)NN = (ab)N.
The normality of N is used in this equation. Because of the normality of N, the left cosets and right cosets of N in G are equal, and so G/N could be defined as the set of right cosets of N in G. Because the operation is derived from the product of subsets of G, the operation is well-defined (does not depend on the particular choice of representatives), associative, and has identity element N. The inverse of an element aN of G/N is a−1N.
For example, consider the group with addition modulo 6:
- G = {0, 1, 2, 3, 4, 5}.
Let
- N = {0, 3}.
The quotient group is:
- G/N = { aN : a ∈ G } = { a{0, 3} : a ∈ {0, 1, 2, 3, 4, 5} } =
- { 0{0, 3}, 1{0, 3}, 2{0, 3}, 3{0, 3}, 4{0, 3}, 5{0, 3} } =
- { {(0+0) mod 6, (0+3) mod 6}, {(1+0) mod 6, (1+3) mod 6},
- {(2+0) mod 6, (2+3) mod 6}, {(3+0) mod 6, (3+3) mod 6},
- {(4+0) mod 6, (4+3) mod 6}, {(5+0) mod 6, (5+3) mod 6} } =
- { {0, 3}, {1, 4}, {2, 5}, {3, 0}, {4, 1}, {5, 2} } =
- { {0, 3}, {1, 4}, {2, 5}, {0, 3}, {1, 4}, {2, 5} } =
- { {0, 3}, {1, 4}, {2, 5} }.
The basic argument above is still valid if G/N is defined to be the set of all right cosets.
Read more about this topic: Quotient Group
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