6 Cosets & Factor Groups


 Emma Pierce
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1 6 Cosets & Factor Groups The course becomes markedly more abstract at this point. Our primary goal is to break apart a group into subsets such that the set of subsets inherits a natural group structure. This sounds strange, but it is precisely what you ve already encountered when constructing a natural addition on the set of remainders modulo n: in Z 6, when we write = 2, we really mean [3] + [5] = [2], where [3] = {3 + 6n : n Z} is the set of integers whose remainder is 3 when divided by 6. We shall call this set a coset to indicate that it is related to a subgroup of the integers, namely 3Z. That the above addition of sets makes sense and defines a group structure is merely an example of a more general construction. 6.1 Lagrange s Theorem Before properly defining cosets, we state what you should already have hypothesized, given the huge number of examples you ve now seen. Cosets will turn out to provide a simple proof. Review any of the subgroup relations we ve already seen and observe that the order of a subgroup is a divisor of the order of its parent group. This is a general result. Theorem 6.1 (Lagrange). Suppose that G is a finite group. The order of any subgroup G divides the order of G. Otherwise said G = G We have already proved the special case for subgroups of cyclic groups: 1 If G is a cyclic group of order n, then, for every divisor d of n, G has exactly one subgroup of order d. More precisely, if G = g has order n, then g k = Cd where d = n gcd(n,k) g k = g l gcd(n, k) = gcd(n, l) We shall prove the full theorem shortly. For such a simple result, Lagrange s Theorem is extremely powerful. For instance, you may have observed that there is only one group structure, up to isomorphism, of prime orders 2, 3, 5 and 7. This is also a general result. Corollary 6.2. There is only one group (up to isomorphism) of each prime order p, namely C p. Proof. Let p be prime and suppose that G is a group with G = p. Since p 2, we may choose some element x G \ {e}. Consider the cyclic subgroup generated by x, namely x G. Lagrange = x divides p = x = 1 or p. Since x has at least two elements, its order must therefore be p, from which G = x is cyclic. Finally, recall that there is only one cyclic group of each order up to isomorphism, so G = C p. 1 Lagrange s Theorem is sometimes misremembered as the order of an element divides the order of a group. This is really only a statement about the cyclic subgroups of a group G and is not as general as Lagrangeproper. 1
2 6.2 Cosets and Normal Subgroups The idea for the proof of Lagrange s Theorem is very simple. Given a subgroup G, partition G into several subsets, each with the same cardinality as. We call these subsets the cosets of. Definition 6.3. Let G and choose g G. The subsets of G defined by g := {gh : h } and g := {hg : h } are (respectively) the left and right cosets of containing g. If the left and right cosets of containing g are always equal ( g G, g = g) we say that is a normal subgroup of G, and write G. If G is written additively, the left and right cosets of containing g are g + := {g + h : h } and + g := {h + g : h } Even though our ultimate purpose is to prove Lagrange s Theorem (a statement about finite groups), the concept of cosets is perfectly welldefined for any group. Before seeing some examples, we state a straightforward lemma giving some conditions for identifying normal subgroups. Lemma Every subgroup of an abelian group G is normal. 2. A subgroup is normal in G if and only if g G, h, ghg 1. For nonabelian groups, most subgroups are typically not normal (although see example 3 below). Proof. Part 1. should be obvious. For part 2., note first that G = g G, h, gh g = g G, h, ghg 1 Conversely, ghg 1 = gh g = g g. If this holds for all g G and h, then it certainly applies for g 1, whence g 1 hg = hg g = g g We conclude that g = g for all g G and so is normal in G. Examples 1. Consider the subgroup of Z 12 generated by 4. This subgroup is cyclic of order 3: = 4 = {0, 4, 8} = C 3 Since the group operation is addition, we write cosets additively: for example, the left coset of 4 containing x Z 12 is the subset x + 4 = {x + n : n 4 } = {x, x + 4, x + 8} 2
3 where we might have to reduce x + 4 and x + 8 modulo 12. The distinct left cosets of 4 are as follows: = {0, 4, 8} = = (usually written 4, dropping the zero) = {1, 5, 9} = = = {2, 6, 10} = = = {3, 7, 11} = = There are only four distinct cosets: notice that each has the same number of elements (three) as the subgroup 4, and that Z 12 = = 4. The cosets have partitioned Z 12 into equalsized subsets. This is crucial for the proof of Lagrange s Theorem. Observe also that the right cosets of 4 are the same as the left cosets, in accordance with Lemma 6.4). Observe finally that only one of the cosets is a subgroup, the others are merely subsets of Z Recall the multiplication table for D 3. With a little work, you should be able to compute that the left and right cosets of the subgroup = {e, µ 1 } are as follows: Left cosets Right cosets e = µ 1 = = {e, µ 1 } e = µ 1 = = {e, µ 1 } ρ 1 = µ 3 = {ρ 1, µ 3 } ρ 1 = µ 2 = {ρ 1, µ 2 } ρ 2 = µ 2 = {ρ 2, µ 2 } ρ 2 = µ 3 = {ρ 1, µ 3 } This time the left and right cosets of are different ( is not a normal subgroup of D 3 ), but all cosets still have the same cardinality. 3. Recall how we proved that the alternating subgroup A n S n has cardinality A n = 1 2 S n. Generalizing this approach, it should be clear that, for any α A n and σ S n, we have ασ even σ even σα even Otherwise said, for any σ S n, the cosets of A n contiaining σ are { A n if σ even σa n = A n σ = if σ odd B n where B n is the set of odd permutations in S n. Note that A n is a normal subgroup of S n. 4. The vector space R 2 is an Abelian group under addition. The real line R identified with the xaxis, is a subgroup. The cosets of R in R 2 are all the sets v + R which are horizontal lines. More generally, if W is a subspace of a vector space V, then the cosets v + W form sets parallel to W: only the zero coset W = 0 + W is a subspace. In the picture, W is line through the origin in R 2 ; its cosets comprise all the lines in R 2 parallel to W (including W itself!). W 3
4 The examples should have suggested the following Theorem. Theorem 6.5. The left cosets of partition G. Before reading the proof, look at each of the above examples and convince yourself that the result is satisfied in each case. The strategy is to define an equivalence relation, the equivalence classes of which are precisely the left cosets of. Proof. Define a relation on G by x y x 1 y. We claim that is an equivalence relation. Reflexivity x x since x 1 x = e. Symmetry x y = x 1 y = (x 1 y) 1, since is a subgroup. But then y 1 x = y x. Transitivity If x y and y z then x 1 y and y 1 z. But is closed, whence x 1 z = (x 1 y)(y 1 z) = x z. The equivalence classes of therefore partition G. We claim that these are precisely the left cosets of. Specifically, we claim that x y x and y lie in the same left coset of x = y owever, note that x y x 1 y y x. It follows that y x. By symmetry, x y y 1 x x y. We conclude that x y x = y as required. Aside: Partitions and Subgroups Each of the three conditions that be an equivalence relation corresponds to one of the properties characterizing as a subgroup of G: Reflexivity contains the identity. Symmetry is closed under inverses. Transitivity is closed under the group operation. It is precisely the fact that is a subgroup which guarantees a partition. For example, if is merely the subset = {0, 1} of Z 3 = {0, 1, 2}, then its left cosets are 0 + {0, 1} = {0, 1}, 1 + {0, 1} = {1, 2}, 2 + {0, 1} = {2, 1}, which do not partition Z 3. 4
5 Proof of Lagrange s Theorem. Suppose that G and g G is a fixed element. Then the function φ : g : h gh is a bijection (its inverse is φ 1 : gh g). It follows that every left coset of has the same cardinality as. Therefore G = (number of left cosets of ) whence divides G. You should now revisit the proof that A n = 1 2 S n : it is nothing but a special case of Theorem 6.5 and the proof of Lagrange. The right cosets of a subgroup also partition G although, in general, they do this in a different way to the left cosets: observe the earlier example of the cosets of the subgroup {e, µ 1 } D 3. The proof of Lagrange s Theorem could also be argued with right cosets. 6.3 Indices The proof of Lagrange s Theorem in fact tells us that the number of left and right cosets of in G is identical: both are equal to the quotient G. This leads us to the following definition. Definition 6.6. If G then the index of in G, written (G : ), is the number of left (or right) cosets of in G. Strictly, the index is the cardinality of the set of left cosets: (G : ) = {g : g G} If G is finite (G : ) = G and the index is simple to calculate. It is more difficult to consider the situation when G is infinite, although the concept is welldefined (the cardinalities of the sets of left and right cosets are identical). Theorem 6.7. If K G is a sequence of subgroups then (G : K) = (G : )( : K) If G is a finite group then the result is essentially trivial: (G : K) = G K = G = (G : )( : K) K owever, the Theorem is more general, for it applies also to infinite groups. 2 2 Even when dealing with infinite groups, we will only think about examples where indices are finite so that their multiplication makes sense. If you are comfortable multiplying infinite cardinals then there is no problem removing this restriction and our proof works even in the general case. 5
6 Proof. Given K G choose a single element g i from each left coset of in G and a single element k j from each left coset of K in. Clearly (G : ) = {g i } and ( : K) = {h j } We claim that the left cosets of K in G are precisely the sets (g i h j )K. Certainly each set (g i h j )K is a coset of K in G: our goal is to show that these sets partition G, whence the collection {(g i h j )K} comprises all the left cosets of K in G. The partition conditions are confirmed below: Every g G lies in some left coset of, so g i G such that g g i. But then g 1 i g. Every h lies in some left coset of K in, so h j such that g 1 i g h j K. But then g (g i h j )K so that every g G lies in at least one set (g i h j )K. Suppose y g i h j K g α h β K. Since the left cosets of partition G, we have y g i g α = g α = g i But then g 1 i y h j K h β K = h β = h j similarly, since the left cosets of K in partition. It follows that the sets (g i h j )K are disjoint. Since the left cosets of K in G are given by {(g i h j )K}, it is immediate that 3 (G : K) = {g i h j } = {g i } {h j } = (G : )( : K) Examples 1. Consider G = Z 20 with its subgroups = 2 and K = {10}. Certainly K = {0, 10} = {0, 2, 4,..., 18} G = {0, 1, 2, 3,...} There are two (left) 4 cosets of in G, namely = {0, 2, 4,..., 18} and 1 + = {1, 3, 5,..., 19} whence the index is (G : ) = 2. Indeed, in the language of the proof, we could choose representatives g 1 = 0 and g 2 = 1. Meanwhile, K has five cosets in : K = {0, 10}, 2 + K = {2, 12}, 4 + K = {4, 14}, 6 + K = {6, 16}, 8 + K = {8, 18} so that ( : K) = 5. We could choose representatives h 1 = 0, h 2 = 2, h 3 = 4, h 4 = 6, h 5 = 8. There are ten cosets of K in G, in accordance with (G : K) = (G : )( : K): K = {0, 10}, 1 + K = {1, 11}, 2 + K = {2, 12},..., 9 + K = {9, 19} In the language of the Theorem, these can be written K = (g 1 + h 1 ) + K, 1 + K = (g 2 + h 1 ) + K, 2 + K = (g 1 + h 2 ) + K,......, 8 + K = (g 1 + h 4 ) + K, 9 + K = (g 2 + h 4 ) + K 3 The fact that {g i h j } = {g i } {h j } follows from the second bulletpoint. 4 The example is Abelian so left and right cosets are identical. 6
7 2. We consider the cosets for the sequence of subgroups C 3 S 3 S 4 where C 3 = {e, (123), (132)} and S 3 = {σ S 4 : σ(4) = 4}. The left cosets for C 3 in S 3 are ec 3 = C 3 = {e, (123), (132)} and (12)C 3 = {(12), (23), (13)} reflecting the fact that (S 3 : C 3 ) = 2. In the language of the Theorem, g 0 = e and g 1 = (12). Similarly, the left cosets for S 3 in S 4 are es 3 = S 3 = {e, (123), (132), (12), (23), (13)} (14)S 3 = S 3 = {(14), (1234), (1324), (124), (14)(23), (134)} (24)S 3 = S 3 = {(24), (1423), (1342), (142), (234), (13)(24)} (34)S 3 = S 3 = {(34), (1243), (1432), (12)(34), (243), (143)} with (S 4 : S 3 ) = 4 and h 0 = e, h 1 = (14), h 2 = (24), h 3 = (34). Finally, the left cosets of C 3 in S 4 are ec 3 = C 3 = {e, (123), (132)} (12)eC 3 = (12)C 3 = {(12), (23), (13)} (14)C 3 = {(14), (1234), (1324)} (12)(14)C 3 = (124)C 3 = {(124), (14)(23), (134)} (24)C 3 = {(24), (1423), (1342)} (12)(24)C 3 = (142)C 3 = {(142), (234), (13)(24)} (34)C 3 = {(34), (1243), (1432)} (12)(34)C 3 = {(12)(34), (243), (143)} in accordance with (S 4 : C 3 ) = 8 = (S 4 : S 3 ) = (S 3 : C 3 ). There are patterns everywhere! Aside: Indices in Infinite Groups ere are two classic examples: It is more challenging to think about indices for infinite groups. 1. O 2 (R) is the set of rotations and reflections of the plane (or of the unit circle if you prefer). Its subgroup SO 2 (R) is the set of rotations. Both groups are infinite (indeed uncountable), but we can compute the index: (O n (R) : SO n (R)) = 2 Intuitively this says that the rotations comprise half of group O 2 (R). To prove this, one performs what should be a nowfamiliar trick: the function ( ) 1 0 φ : SO 2 (R) O 2 (R) \ SO 2 (R) : A A 0 1 is a bijection The sets Q and Z are both groups under addition. Their index is countably infinite! (Q : Z) = ℵ 0 To see this, note that p + Z = q + Z p q Z so that there is precisely one coset of Z in Q for every rational number in the interval [0, 1), a denumerable set. 7
8 6.4 Factor Groups The idea of factor groups is very simple, if abstract. Given a subgroup G we may consider the set of left cosets G / = {g : g G} The question for this section is whether the set of left cosets has any interesting structure: that is, can we view G/ as a group in a natural way?5 To see how this might work, let us recall the first two examples of cosets starting on page 2. Turning the set of left cosets into a group cosets Consider Z 12 where = 4 = {0, 4, 8}, with (left), 1 +, 2 +, 3 + The set of left cosets is then / { } Z 12 = {, 1 +, 2 +, 3 + } = {0, 4, 8}, {1, 5, 9}, {2, 6, 10}, {3, 7, 11} There is a nice pattern here: x and y are members of the same coset if and only if x y (mod 4). Moreover, there are four cosets. Can we view Z / 12 as merely the cyclic group Z 4 in disguise? The answer, of course, is yes. To do this, we need to define a binary operation on Z / 12 : we need to see how to combine cosets to create new ones. The obvious thing to do is to use the addition in Z 12 that we already have: thus define (a + ) (b + ) := (a + b) + The binary operation on the set Z / 12 comes naturally from the existing addition on Z 12: we haven t created anything new, just using what we ve been given. There is a potential problem however: the freedom of choice built into the definition. Computing (a + ) (b + ) requires three steps: Choose representatives Make a choice of elements a and b in the respective cosets. Add in the original group Compute a + b Z 12. Take the coset Return the left coset (a + b) +. 5 There are typically many operations one could define on a set which would define a group structure. We want the structure to be natural: this means that it should appear without choice or imposition, and be inherited from the preexisting structure on G. In this way the structure will encode some information about G itself. 8
9 If is to make any sense at all, then the result (a + b) + must be independent of the choices we made in the first step. 6 In this case there is no problem, as you can tediously check for yourself. For example, to compute (1 + ) (2 + ) there are nine possibilities: = 3 = (1 + ) (2 + ) = = 7 = (1 + ) (2 + ) = 7 + = = 11 = (1 + ) (2 + ) = 11 + = = 7 = (1 + ) (2 + ) = 7 + = = 11 = (1 + ) (2 + ) = 11 + = = 15 = (1 + ) (2 + ) = 15 + = = 11 = (1 + ) (2 + ) = 11 + = = 15 = (1 + ) (2 + ) = 15 + = = 19 = (1 + ) (2 + ) = 19 + = 3 + Thankfully all nine results are the same! The other possibilities for adding cosets can be similarly checked (we ll do this in general in a moment). It is not hard to produce a table for the binary operation on the set Z / 12 : This table should look very familiar: if you delete all the expressions (so that = 0 + becomes 0), we recover precisely the Cayley table for the cyclic group Z 4. Indeed we have proved the following: Proposition 6.8. The set of left cosets Z / 12 = {, 1 +, 2 +, 3 + } forms a group under the operation, which is moreover isomorphic to Z 4. We will shortly refer to this as a factor group. It would be very nice if this sort of behavior were universal. Unfortunately it isn t. When cosets fail to form a group Let us repeat the process with the subgroup = {e, µ 1 } D 3. Recall that its left cosets are e = µ 1 = = {e, µ 1 }, ρ 1 = µ 3 = {ρ 1, µ 3 }, ρ 2 = µ 2 = {ρ 2, µ 2 } / In the same way as above, we define the natural operation on the set D 3 of left cosets:7 a b := (ab) 6 Strictly speaking we require that : Z / 12 Z / 12 Z / 12 be a welldefined function. 7 Again, the operation on the set of left cosets comes directly from the group operation on D 3. 9
10 This time there is a problem. Suppose we want to compute ρ 1 ρ 1. There are four choices: 8 ρ 1 ρ 1 = ρ 2 1 = ρ 2 ρ 1 µ 3 = ρ 1 µ 3 = µ 2 = ρ 2 µ 3 ρ 1 = µ 3 ρ 1 = µ 1 = µ 3 µ 3 = µ 2 3 = e = Exercising the freedom of choice in the definition of leads to different outcomes: it follows that is not welldefined. There is nothing stopping us from defining a group structure on D / 3, but any such definition must involve a choice on our part: we would be imposing structure rather that observing structure that occured naturally. Welldefinition of the natural group structure The above discussion can be generalized. Clearly some subgroups G behave better than others when considering the set G/ of left cosets. ow can we tell which subgroups? Let be a subgroup of G and define the natural operation on G/ : a b := (ab) This is welldefined if and only if a, b G, x a, y b, we have a b = x y Let us trace through what this means for the subgroup. x a h such that x = ah The operation on G/ is therefore welldefined if and only if a, b G, h 1, h 2, (ah 1 bh 2 ) = (ab) a, b G, h, (ahb) = (ab) (since h 2 = for all h 2 ) a, b G, h, (ab) 1 (ahb) b G, h, b 1 hb G We ve therefore proved the critical part of the following: (recall Lemma 6.4) Theorem 6.9. Suppose that is a subgroup of G. The set of (left) cosets G/ natural operation forms a group under the a b := (ab) if and only if is a normal subgroup of G. 8 Write these as cycles if you re having trouble computing: e.g. ρ 1 = (123), µ 1 = (23), etc. 10
11 Proof. The above discussion proves the direction: if G/ forms a group then the operation is certainly welldefined, whence is normal. For the converse, we check the group axioms. Closure By the above discussion, G ( = the natural operation is welldefined. Certainly a b = (ab) is a coset, whence G / ), is closed. Associativity a (b c) = a (bc) = a(bc). Similarly (a b) c = (ab)c. By the associativity of G these are identical. Identity e a = (ea) = a therefore e = is the identity. Inverse a 1 a = (a 1 a) = e =, therefore (a) 1 = a 1. Definition If is a normal subgroup of G, then G/ is called a factor group. Warning! Because the group structure on a factor group G/ comes naturally from the group structure on G, we typically use the same notation for the operation. Thus if (G, +) is a group, we also use + for the operation on G/ ( / ). Similarly (G, ) will have factor groups G, written multiplicatively. Make sure you know to which group an operation refers! The symbols, in the above examples were used only to help you keep the operations separate. 6.5 Factor Groups: Examples and Calculations There are many nice examples of factor groups. The main question we are concerned with is how to identify a given factor group G/ : by this we mean that we want to find a wellunderstood group K which is isomorphic to G/. We start with a general example which is important enough to merits a full discussion. This should, however, be a revision of material from a previous class. Factor Groups of Infinite Cyclic Groups: The Group Z m For each integer m, its integer multiples mz = m form a subgroup of Z. Since Z is abelian, mz is a normal subgroup. Observe that x + mz = y + mz x y mz x y mod m whence there is precisely one coset for each remainder modulo m. We may therefore write the cosets of mz as Z / = {0 + mz, 1 + mz,..., (m 1) + mz} mz Addition in the factor group is natural: ( ) ( ) x + mz + y + mz = (x + y) + mz This looks suspiciously like addition modulo m! Indeed we have the following: 11
12 Theorem µ : Z/ mz Z m : x + mz x (mod m) is an isomorphism. Proof. We ve already seen most of the relevant calculations! Welldefinition and Injectivity Note that x + mz = y + mz x y mz x y 0 (mod m) x = y (in Z m ) µ(x + mz) = µ(y + mz) Surjectivity Given any x Z m, we have x = µ(x + mz). omomorphism For all x + mz, y + mz Z/, we have mz ( ) ( ) µ (x + mz) + (y + mz) = µ (x + y) + mz = x + y (mod m) = µ ( x + mz ) + µ ( y + mz ) (mod m) By this argument, algebraists often take the factor group Z/ mz to be the definition of the group Z m. Factor Groups of Finite Cyclic Groups We ve already seen the example Z / 12 4 = Z 4 (Proposition 6.8). We can, in fact, identify all factor groups of finite cyclic groups. Recall that Z n has a single subgroup of order n d for each divisor d of n, namely9 { ( n ) } d = 0, d, 2d,..., d 1 d Theorem If d n, then µ : Z / n d Z d : x + d x (mod d) is an isomorphism. Try proving this yourself: it should be very similar to the proof of Theorem Example 5 = {0, 5, 10, 15} Z 20 has factor group Z 20 = {0 + 5, 1 + 5, 2 + 5, 3 + 5, } / 5 This is isomorphic to Z 5 under the isomorphism µ : Z / 20 5 Z 5 : x + 5 x (mod 5) 9 Recall that s = d gcd(s, n) = gcd(d, n) = d. 12
13 Factor Groups of Finite Abelian Groups If G is a finite abelian group with subgroup, then G/ is also a finite abelian group. According to the Fundamental Theorem of Finitely Generated Abelian Groups, the means that G / = Z m1 Z mk for some m 1,..., m k N which satisfy m 1 m k = (G : ) = G. Our goal in the following examples is to identify G/ by finding the relevant m k. Examples 1. Identify the factor group G/ = (Z 4 Z 8 ) / in terms of the Fundamental Theorem of (0, 1) Finitely Generated Abelian Groups. First consider the elements of the cyclic subgroup: = (0, 1) = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7)} This has order 8 whence the number of cosets is index of in G, namely (G : ) = We can easily compute the cosets: recall that (x, y) + = (v, w) + (x, y) (v, w) = (x v, y w) x = v It follows that there is a unique element in each coset of the form (x, 0), where x Z 4. Indeed the factor group may be written G / { } =, (1, 0) +, (2, 0) +, (3, 0) + The factor group is abelian of order 4. We therefore have two possibilities: G / = Z 4 or G / = Z2 Z 2. Which is it? A straightforward answer comes by observing that the factor group is generated by the coset (1, 0) +, indeed (1, 0) + = {, (1, 0) +, (2, 0) +, (3, 0) + } = G/ The factor group is cyclic and is therefore isomorphic to Z 4. If you want to be more explicit, observe that the function ψ : Z 4 G/ : x (x, 0) + is an isomorphism. 13
14 2. We can do something similar for G/ = (Z 4 Z 8 ) /. This time the cyclic subgroup (0, 2) = {0, 0), (0, 2), (0, 4), (0, 6)} has order 4. As with the previous example, the elements of of the cyclic subgroup only influce the second factor Z 8 of G, and we therefore expect to see (Z 4 Z 8 ) / (0, 2) = Z 4 Z / 8 2 = Z 4 Z 2 Indeed this is what we find, for the function ψ : Z 4 Z 2 G/ : (x, y) (x, y) + is an isomorphism. If the above seems too, fast, try counting cosets. Clearly G/ = = 8, whence the quotient group is abelian of order 8. There are three nonisomorphic possibilities: G / = Z 8, Z 4 Z 2, Z 2 Z 2 Z 2 To distinguish these, consider the possible orders of elements in each group: Observe: Group Possible orders of elements Z 8 1, 2, 4, 8 Z 4 Z 2 1, 2, 4 Z 2 Z 2 Z 2 1, 2 (1, 0) + has order 4, since (1, 0) + = {, (1, 0) +, (2, 0) +, (3, 0) + } All elements of the factor group have order dividing 4: 4 ( (x, y) + ) = (4x, 4y) + = (0, 4y) + = 2y ( (0, 2) + ) = It follows that the factor group must be isomorphic to Z 4 Z This time we identify 10 G/ = (Z 4 Z 8 ) /. As ever, first compute the subgroup: (2, 4) = {(0, 0), (2, 4)} has order 2. The factor group is therefore abelian of order gives five distinct options for the factor group, namely = 16. The Fundamental Theorem Z 16, Z 2 Z 8, Z 4 Z 4, Z 2 Z 2 Z 4, Z 2 Z 2 Z 2 Z 2 A little scratch work allows us to identify which is correct: 10 The previous examples may have lulled you into a false sense of security: the answer to this question is not Z 4 / 2 Z / 8 4 = Z 2 Z 2 Indeed by counting the sixteen cosets, we see immediately see that this is impossible! 14
15 If x = 2n is even, then (x, y) + = (2n, y) + = (0, y 4n) +. If x = 2n + 1 is odd, then (x, y) + = 1, y 4n) +. It follows that there is precisely one representative of each coset whose first entry is either 0 or 1, whence the following sixteen elements lie in distinct cosets of (2, 4) : (0, 0), (0, 1),..., (0, 7), (1, 0),..., (1, 7) It now seems reasonable to conjecture that the factor group is isomorphic to Z 2 Z 8. ere are two possible ways to prove this: (a) Observe that the coset (0, 1) + has order 8 in the factor group. This narrows our choice to Z 16 or Z 2 Z 8. Now check that all cosets have order at most 8: ( ) 8 (x, y) + = (8i, 8j) + = (0, 0) + There are no elements of order 16 whence we can rule out Z 16 as a candidate. The only possibility remaining is Z 2 Z 8. (b) We can define an explicit isomorphism: ψ : Z 2 Z 8 Z 4 Z 8 : (x, y) (x mod 2, y 4x) + (2, 4) / (2, 4) We leave it as an exercise to check that ψ is a welldefined isomorphism: it requires some creativity to invent such a function from nothing! In practice, method (a) is generally easier for us. Other Examples There are many other examples of factor groups. Often these have to be considered individually. 1. Let = {2πn : n Z}. This is a subgroup of the abelian group of (R, +). Can we identify the factor group R/? It should be clear that in any given coset there is a unique element x such that 0 x < 2π (this is like taking the remainder of x modulo 2π!). It follows that R / = {x + : x [0, 2π)} Indeed, you can check that the function ψ : R/ S1 : x + e ix is a welldefined group isomorphism! The factor group construction therefore corresponds to wrapping the real line infinitely many times around a circle of circumference 2π. 15
16 2. Recall the description of the alternating group A 4. It can be (tediously) checked that V = {e, (12)(34), (13)(24), (14)(23)} is a normal subgroup of A 4 which is moreover isomorphic to the Klein 4group. In this case, the factor group A / 4 V has order 12 4 = 3 and so must be isomorphic to Z 3: can you find an explicit isomorphism? 3. A similar analysis may be performed to see that S / 4 = S 3. See the homework. V Aside: An factor group of an infinite cyclic group As a final example we do something far more difficult: nothing this tricky is examinable! Identify G/ = (Z 10 Z 6 Z) / (4, 2, 3). The challenge is that both G and are infinite, whence we cannot simply use the index formula to count the number of cosets. ere is a way round the problem. Let z = 3q + r be the result of the division algorithm applied to z Z, where r = 0, 1 or 2. Then (x, y, z) + = (x, y, z) q(4, 2, 3) + = (x 4q, y 2q, r) + It follows that in every coset (x, y, z) +, there is precisely one element (x, y, z) with z = 0, 1 or 2. The elements (x, y, 0) where x Z 10 and y Z 6 all lie in different cosets: if two such were in the same coset, then (x 1, y 1, 0) (x 2, y 2, 0) (x 1 x 2, y 1 y 2, 0) x 1 x 2 (mod 10) and y 1 y 2 (mod 6) Similarly all the elements (x, y, 1) and (x, y, 2) lie in different cosets. It follows that the elements (x, y, 0), (x, y, 1) and (x, y, 2) are representatives of different cosets, and that all cosets have one such representative. There are therefore = 180 distinct cosets whence the factor group G/ is abelian of order 180. The prime decomposition of 180 is : there are, up to isomorphism, four abelian groups of order 180, namely Z 4 Z 9 Z 5 = Z180 Z 2 Z 2 Z 9 Z 5 = Z2 Z 90 Z 4 Z 3 Z 3 Z 5 = Z3 Z 60 Z 2 Z 2 Z 3 Z 3 Z 5 = Z6 Z 30 ow do we distinguish between these? As before, we look for elements of a particular order (lots of scratch work may be required in general!). 16
17 Compute the order of the coset (1, 1, 1) +. Certainly its order must be divisible by 3, since the third entry of n(1, 1, 1) = (n, n, n) requires 3 n. Thus let n = 3k. Now 3k(1, 1, 1) + = (3k, 3k, 3k) + = (3k, 3k, 3k) k(4, 2, 3) + (since (4, 2, 3) ) = ( k, k, 0) + = 10 k and 6 k 30 k 90 n Since G/ contains an element of order 90, our choices are narrowed to Z 180 and Z 2 Z 90. We show that every element of G/ has order dividing 90: 90 ( (x, y, z) + ) = (90x, 90y, 90z) 30z(4, 2, 3) + = (90x 120z, 90y 60z, 0) + = (0, 0, 0) + We conclude that G/ = Z2 Z 90. Phew! You can check (it s hard!) that the function ψ : Z 2 Z 90 (Z 10 Z 6 Z) / : (x, y) (y, 3x + y, y) + (4, 2, 3) is an isomorphism, though you d have to be truly inspired to conjecture this out of thin air! 17
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