# 2MA105 Algebraic Structures I

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1 2MA105 Algebraic Structures I Per-Anders Svensson Lecture 7

2 Cosets once again Factor Groups Some Properties of Factor Groups Homomorphisms November 28, (20)

3 Cosets once again Let G be a group, H a subgroup of G, and x an element of G. We recall that the left coset of H represented by x is the set xh = {xh h H }. The coset xh is obtained by multiplying each element of H from the left by x. Similarly the right coset of H represented by x is the set Hx = {hx h H }. Two different elements may represent the same coset; we have xh = yh x 1 y H and Hx = Hy yx 1 H. If additive notation is used in G, we write x + H = {x + h h H } for the (left) coset represented by x. With this notation, two cosets x + H and y + H are equal, if and only if x y H. November 28, (20)

4 Factor Groups Example The left and right cosets of the subgroup 3Z of Z are equal (since Z is Abelian). The cosets of 3Z are exactly the same as the residue classes modulo 3: 0 + 3Z = [0] = {..., 6, 3, 0, 3, 6,... } 1 + 3Z = [1] = {..., 5, 2, 1, 4, 7,... } 2 + 3Z = [2] = {..., 4, 1, 2, 5, 8,... } In general, the cosets of the subgroup nz of Z are exactly the residue classes modulo n. In other words, the set of all cosets of nz is Z n = {[0], [1],..., [n 1]}. Here we have for each r = 0, 1,..., n 1, that [r] is a set; it is the set of all integers [r] = {r + nk k Z} that gives the remainder r when divided by n. November 28, (20)

5 As we previously have seen, Z n = {[0], [1],..., [n 1]} can be made into a group by defining the sum of two residue classes modulo n as [a] [b] = [a + b]. The sum of the residue class modulo n that contains a and the residue class modulo n that contains b is equal to the residue class modulo n that contains a + b. Since a residue class module n is the same as a (left) coset of nz, we could as well say: The sum of the coset of nz represented by a and the coset of nz represented by b is equal to the coset of nz represented by a + b. Or in other words: (a + nz) + (b + nz) = (a + b) + nz. November 28, (20)

6 Can we generalize this? Given a group G and a subgroup H, we want to define a group structure on the set of all left cosets (or right cosets), according to the following rule: The product of the left (right) coset of H represented by x and the left (right) coset of H represented by y is equal to the left (right) coset of H represented by xy. Using mathematical notation: (xh )(yh ) = (xy)h (Hx)(Hy) = H (xy) for left cosets for right cosets. November 28, (20)

7 Example In an example from an earlier lecture, we computed all left cosets of the subgroup H = {ε, µ 3 } in S 3. These are εh = H = {ε, µ 3 }, µ 1 H = {µ 1 ε, µ 1 µ 3 } = {µ 1, ρ 2 }, µ 2 H = {µ 2 ε, µ 2 µ 3 } = {µ 2, ρ 3 }. We try to multiply the two left cosets µ 1 H and µ 2 H : (µ 1 H )(µ 2 H ) = (µ 1 µ 2 )H = ρ 3 H = µ 2 H. Now µ 1 H = ρ 2 H (since µ 1 and ρ 2 belong to the same left coset). Hence we should also have (ρ 2 H )(µ 2 H ) = ρ 3 H. Do we...? No, it turns out that (ρ 2 H )(µ 2 H ) = µ 3 H = H. Multiplication of right cosets of H does not work either (exercise!) However, it will function properly if we try to multiply left cosets of the subgroup A 3 = {ε, ρ 2, ρ 3 } (exercise!) November 28, (20)

8 What is wrong with the subgroup H in the example above, compared to the subgroup A 3? Definition (Normal subgroup) A subgroup N of a group G is called normal, if gn = Ng for all g G, i.e. if the set of all left cosets in G coincides with the set of all right cosets. We sometimes write N G to signify that N is a normal subgroup of G. Example The reason that multiplication of cosets of A 3 works (but not cosets of H = {ε, µ 3 }), in the example above, is due to A 3 being a normal subgroup of S 3, but H being not. Is there a simple way do determine whether a subgroup is normal or not? Theorem Let N be a subgroup of G. Then N G, if and only if gng 1 N for all g G, n N. November 28, (20)

9 Example Let GL(n, R) be the general linear group (of all invertible n n matrices with real elements). Then SL(n, R) = {A GL(n, R) det A = 1} a normal subgroup of GL(n, R) (the so-called special linear group). We show that we at least have SL(n, R) GL(n, R) in the usual way (checking closure, identity and inverse). To show that SL(n, R) GL(n, R), take any A SL(n, R) and B GL(n, R). Then det A = 1 and det B 0. We want to show that BAB 1 SL(n, R), i.e. that det(bab 1 ) = 1. The laws of arithmetics for determinants yields as claimed. det(bab 1 ) = det B det A 1 det B Exercise Show that if G is a group, then Z (G) G, where = det A = 1, Z (G) = {z G zg = gz for all g G}. November 28, (20)

10 Theorem Let G be a group and N a subgroup of G. Furthermore, let L N denote the set of all left cosets of N. Then multiplication of left coset according to (gn )(hn ) = ghn (1) is a well-defined binary operation on L N (i.e. the result of a product does not depend on the choice of representatives of the cosets), if and only if N G. Corollary Suppose G is a group and N is a normal subgroup of G. Then L N is a group with respect to the binary operation (1). Definition (Factor Group) The group L N in the corollary above is the so-called factor group of G modulo N. It is denoted G/N. Remark The identity element of G/N is the coset that contains the identity element of G, so 1 G/N = 1 G N = N. The inverse of an element an G/N is the coset that contains a 1, whence (an ) 1 = a 1 N. November 28, (20)

11 Example Consider Z 7, the group of units modulo 7. Then H = 6 = {1, 6} is normal subgroup of Z 7, of order 2. The factor group Z 7/H contains, besides H (which is the identity element of the factor group), the cosets 2H = {2, 5} and 3H = {3, 4}. A Cayley table of Z 7/H is: The group is isomorphic to Z 3. H 2H 3H H H 2H 3H 2H 2H 3H H 3H 3H H 2H November 28, (20)

12 Example Write down a Cayley table of the factor group Z 16/H, where H = 9. Solution. We have 9 2 = 1 in Z 16, whence H = {1, 9}. The cosets of H (i.e. the elements of Z 16/H ) are H, 3H = {3, 11}, 5H = {5, 13}, and 7H = {7, 15}. A Cayley table may look like H 3H 5H 7H H H 3H 5H 7H 3H 3H H 7H 5H 5H 5H 7H H 3H 7H 7H 5H 3H H The group is clearly Abelian. According to the Fundamental Theorem of Finite Abelian Groups, there are two groups of order 4: Z 4 and Z 2 Z 2. Which one of these is Z 16/ 9 isomorphic to? November 28, (20)

13 Some Properties of Factor Groups Theorem Let G be a group and N a normal subgroup of G. Then the following holds: (i) If G is Abelian, then G/N is Abelian. (ii) If G is cyclic, then G/N is cyclic. (iii) The order of an element an G/N equals the smallest positive integer m such that a m N, if such an m exists. Otherwise the order of an is infinite. (iv) If G is finite, then G/N is also finite and G/N = G / N. November 28, (20)

14 Example Classify the factor group G = (Z 9 Z 8 )/ (6, 4), according to the Fundamental Theorem of Finite Abelian Groups. Solution. It is clear that G is Abelian, since Z 9 Z 8 is. Furthermore o(6) = 9/ gcd(9, 6) = 9/3 = 3 in Z 9, and o(4) = 8/ gcd(8, 4) = 8/4 = 2 in Z 8, which yields (6, 4) = lcm(3, 2) = 6. Since Z 9 Z 8 = 9 8 = 72, the order of G = (Z 9 Z 8 )/ (6, 4) is 72/6 = 12. Since 12 = 2 2 3, the Fundamental Theorem of Finite Abelian Groups tells us that G could be isomorphic either to Z 4 Z 3 or to Z 2 Z 2 Z 3. The structural difference between these two groups is that Z 4 Z 3 is cyclic, but Z 2 Z 2 Z 3 is not. Since Z 9 Z 8 is cyclic (gcd(9, 8) = 1), every factor group constructed from this group must be cyclic as well. Hence (Z 9 Z 8 )/ (6, 4) Z 4 Z 3 Z 12. November 28, (20)

15 Homomorphisms We recall that an isomorphism between two groups G and H is a bijective mapping φ: G H such that φ(ab) = φ(a)φ(b) for all a, b G. If φ is an isomorphism, then G and H are isomorphic; essentially the same group. For example Z 4 = {0, 1, 2, 3} (addition modulo 4) is isomorphic to Z 5 = {1, 2, 3, 4} (multiplication modulo 5), since the mapping φ: Z 4 Z 5, defined by the table below, is bijective and fulfills φ(a + b) = φ(a)φ(b) for all a, b Z 4 : a φ(a) November 28, (20)

16 We obtain a more general class of mappings, if we do not require φ to be bijective: Definition (Homomorphism) Let G and H be groups. A mapping φ: G H such that φ(ab) = φ(a)φ(b) for all a, b G, is called a homomorphism. Example Let R be the multiplicative group of all non-zero real numbers, and C the multiplicative group of all non-zero complex numbers. Then φ: C R defined by φ(z) = z, for all z C, is a homomorphism, since φ(z 1 z 2 ) = z 1 z 2 = z 1 z 2 = φ(z 1 )φ(z 2 ). But φ is not an isomorphism. Why? November 28, (20)

17 Lemma If φ: G H is a homomorphism between groups, then 1. φ(1 G ) = 1 H 2. φ(a 1 ) = φ(a) 1 3. φ(a n ) = φ(a) n for all n Z. Loosely speaking: Identity elements are mapped onto identity elements, inverses onto inverses, and powers onto powers. There are two important subgroups connected to a homomorphism between groups. Theorem Let φ: G H be a homomorphism between groups. Then 1. im φ = φ[g] = {φ(g) g G} is a subgroup of H 2. ker φ = {g G φ(g) = 1 H } is a subgroup of G. Definition (Image and Kernel of a Homomorphism) Let G and H be groups, and φ: G H a homomorphism. The subgroup im φ of H is called the image of φ, while the subgroup ker φ of G goes by the name of the kernel of φ. November 28, (20)

18 Example In an earlier example we considered the homomorphism φ: C R defined by φ(z) = z for each z C. Here we have im φ = { z z C } = R + and (since the identity element of R is 1) ker φ = {z C z = 1}. Theorem Let φ: G H be a homomorphism between groups. Then ker φ is a normal subgroup of G. Proof. We have to show that gng 1 ker φ, whenever g G and n ker φ: φ(gng 1 ) = φ(g)φ(n)φ(g 1 ) = φ(g)1 H φ(g) 1 = 1 H. Thus, for a given homomorphism φ: G H between groups, we may in a natural way construct a factor group G/ ker φ. November 28, (20)

19 Example Define φ: Z Z n as φ(a) = [a] for all a Z. (We map each a Z onto its residue class [a] modulo n.) Then φ is a homomorphism, since Furthermore φ(a + b) = [a + b] = [a] [b] = φ(a) φ(b). ker φ = {a Z φ(a) = [0]} = {a Z a = nk for some k Z} = nz, i.e. the corresponding factor group Z/ ker φ is Z/nZ. We can thus, given any homomorphism of groups, construct a factor group, using the procedure above. As we soon will see, it is also possible to go the other way around: Given a factor group, we may construct a homomorphism. November 28, (20)

20 Theorem Let G be a group and suppose N G. Then there is a group H and a homomorphism φ: G H such that ker φ = N. Proof (Sketch). Let H = G/N and define ψ N : G H as ψ N (a) = an for all a G (map each a onto the (left) coset of N that contains a). Then ψ N is a homomorphism, having the kernel N. Definition (Canonical Homomorphism) Let G be a group and N G. Then the homomorphism ψ N : G G/N above is called the canonical homomorphism. Example The canonical homomorphism ψ nz : Z Z/nZ is defined as φ(a) = a + nz for all a Z. November 28, (20)

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