2.7. Quotient Groups and Homomorphism Theorems
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“bookmt” — 2006/8/8 — 12:58 — page 132 — #144
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2. BASIC THEORY OF GROUPS
Note that the center of a group is related to the notion of conjugacy in the following way: The center consists of all elements whose conjugacy class is a singleton. That is, g 2 Z.G/ , the conjugacy class of g is fgg.
Exercises 2.6 2.6.1. Consider any surjective map f from a set X onto another set Y . We can define a relation on X by x1 x2 if f .x1/ D f .x2/. Check that this is an equivalence relation. Show that the associated partition of X is the partition into “fibers” f 1.y/ for y 2 Y .
The next several exercises concern conjugacy classes in a group.
2.6.2. Show that conjugacy of group elements is an equivalence relation.
2.6.3. What are the conjugacy classes in S3?
2.6.4. What are the conjugacy classes in the symmetry group of the square D4?
2.6.5. What are the conjugacy classes in the dihedral group D5?
2.6.6. Show that a subgroup is normal if, and only if, it is a union of conjugacy classes.
2.7. Quotient Groups and Homomorphism Theorems
Consider the permutation group Sn with its normal subgroup of even permutations. For the moment write E for the subgroup of even permutations and O for the coset O D .12/E D E.12/ consisting of odd permutations. The subgroup E is the kernel of the sign homomorphism W Sn !
f1; 1g.
Since the product of two permutations is even if, and only if, both are even or both are odd, we have the following multiplication table for the two cosets of E:
E
O
E
E
O
O
O
E
The products are taken in the sense mentioned previously; namely the product of two even permutations or two odd permutations is even, and the product of an even permutation with an odd permutation is odd. Thus the multiplication on the cosets of E reproduces the multiplication on the group f1; 1g.
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“bookmt” — 2006/8/8 — 12:58 — page 133 — #145
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2.7. QUOTIENT GROUPS AND HOMOMORPHISM THEOREMS
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This is a general phenomenon: If N is a normal subgroup of a group G, then the set G=N of left cosets of a N in G has the structure of a group.
The Quotient Group Construction Theorem 2.7.1. Let N be a normal subgroup of a group G. The set of cosets G=N has a unique product that makes G=N a group and that makes the quotient map W G ! G=N a group homomorphism.
Proof. Let A and B be elements of G=N (i.e., A and B are left cosets of N in G). Let a 2 A and b 2 B (so A D aN and B D bN ). We would like to define the product AB to be the left coset containing ab, that is, .aN /.bN / D abN: But we have to check that this makes sense (i.e., that the result is independent of the choice of a 2 A and of b 2 B). So let a0 be another element of aN and b0 another element of bN . We need to check that abN D a0b0N , or, equivalently, that .ab/ 1.a0b0/ 2 N . We have .ab/ 1.a0b0/ D b 1a 1a0b0 D b 1a 1a0.bb 1/b0 D .b 1a 1a0b/.b 1b0/: Since aN D a0N , and bN D b0N , we have a 1a0 2 N and b 1b0 2 N .
Since N is normal, b 1.a 1a0/b 2 N . Therefore, the final expression is a product of two elements of N , so is in N . This completes the verification that the definition of the product on G=H makes sense.
The associativity of the product on G=N follows from repeated use of the definition of the product, and the associativity of the product on G; namely .aN bN /cN D abNcN D .ab/cN D a.bc/N D aN bcN D aN.bNcN /: It is clear that N itself serves as the identity for this multiplication and that a 1N is the inverse of aN . Thus G=N with this multiplication is a group.
Furthermore, is a homomorphism because .ab/ D abN D aN bN D .a/.b/:
The uniqueness of the product follows simply from the surjectivity of : in order for to be a homomorphism, it is necessary that aN bN D abN .
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“bookmt” — 2006/8/8 — 12:58 — page 134 — #146
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2. BASIC THEORY OF GROUPS The group G=N is called the quotient group of G by N . The map W G ! G=N is called the quotient homomorphism. Another approach to defining the product in G=N is developed in Exercise 2.7.2.
Example 2.7.2. (Finite cyclic groups as quotients of Z). The construction of Zn in Section 1.7 is an example of the quotient group construction.
The (normal) subgroup in the construction is nZ D f`n W ` 2 Zg. The cosets of nZ in Z are of the form k C nZ D Œk; the distinct cosets are Œ0 D nZ; Œ1 D 1 C nZ; : : : ; Œn
1 D n 1 C nZ. The product (sum) of two cosets is Œa C Œb D Œa C b. So the group we called Zn is precisely Z=nZ. The quotient homomorphism Z ! Zn is given by k 7! Œk.
Example 2.7.3. Now consider a cyclic group G of order n with generator a. There is a homomorphism ' W Z ! G of Z onto G defined by '.k/ D ak. The kernel of this homomorphism is precisely all multiples of n, the order of a; ker.'/ D nZ. I claim that ' “induces” an isomorphism Q ' W Zn ! G, defined by Q '.Œk/ D ak D '.k/. It is necessary to check that this makes sense (i.e., that Q ' is well defined) because we have attempted to define the value of Q ' on a coset Œk in terms of a particular representative of the coset. Would we get the same result if we took another representative, say k C 17n instead of k? In fact, we would get the same answer: If Œa D Œb, then a b 2 nZ D ker.'/, and, therefore, '.a/ '.b/ D '.a b/ D 0.
Thus '.a/ D '.b/. This shows that the map Q ' is well defined.
Next we have to check the homomorphism property of Q '. This prop erty is valid because Q'.ŒaŒb/ D Q'.Œab/ D '.ab/ D '.a/'.b/ D Q'.Œa/ Q'.Œb/: The homomorphism Q ' has the same range as ', so it is surjective. It also has trivial kernel: If Q '.Œk/ D 0, then '.k/ D 0, so k 2 nZ D Œ0, so Œk D Œ0. Thus Q ' is an isomorphism.
Example 2.7.4. Take the additive abelian group R as G and the subgroup Z as N . Since R is abelian, all of its subgroups are normal, and, in particular, Z is a normal subgroup.
The cosets of Z in R were considered in Exercise 2.5.11, where you were asked to verify that the cosets are parameterized by the set of real numbers t such that 0 t