ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH 1. Homomorphisms and isomorphisms between groups. Definition 1.1. Let G, H be groups. We say that a function f : G H is a homomorphism of groups (also sometimes called a morphism of groups, a map of groups, or sometimes just a homomorphism, morphism, or map) if the two following axioms are satisfied: Unitality : f (e) = e, that is, f sends the unit element of G to the unit element of H, and Commuting with the group operation : µ( f (x), f (y)) = f (µ(x, y)) for all x, y G. Example 1.2. Let Z/2Z and Z/4Z be the cyclic groups with 2 and 4 elements, respectively. I will write these groups additively, so that 0 is the unit element of each of them, Z/2Z has two elements 0, 1 and 1 + 1 = 0, and Z/4Z has four elements 0, 1, 2, 3, with the group operation being addition modulo 4, so that (for example) 2 + 3 = 1. Let f : Z/2Z Z/4Z be the function given by f (0) = 0 and f (1) = 2. I claim that f is a group homomorphism. Clearly the unitality axiom is satisfied, since f (0) = 0. So the commuting with the group operation axiom has to be checked now. Let s do this in the Date: September 2016. 1
2 ANDREW SALCH most explicit way: f (0 + 0) = f (0) = 0 = 0 + 0 f (0 + 1) = f (1) = f (0) + f (0), = 2 = 0 + 2 f (1 + 0) = f (1) = f (0) + f (1), = 2 = 2 + 0 f (1 + 1) = f (0) = f (1) + f (0), = 0 = 2 + 2 = f (1) + f (1). So the commuting with the group operation axiom is satisfied for each pair of elements x, y Z/2Z. The explicit computation in Example 1.2 wasn t difficult, but you can tell already that this isn t a very good way to check that a function between groups is a group homomorphism: there were four equations f (x + y) = f (x) + f (y) that we had to verify for our function with domain Z/2Z, but if you were considering a function f : G H and G is a group with n elements, then you would need to verify n 2 equations to check that f is a group homomorphism. Clearly there ought to be better ways to check that a function is a group homomorphism! Indeed there are: if you find generators and relations for the group G, then to produce a group homomorphism f : G H, you only need to define f (g) for each generator g of G, and then to check that f respects all of the relations between generators in G. We haven t seen a definition of generators and relations in class yet, but this will come very, very soon. Proposition 1.3. Suppose that F, G, H are groups, and f : F G and g : G H are group homomorphism. Then the composite function g f : F H is also a group homomorphism. Proof. To ease the notation, I will use multiplicative notation for F, G, and H. In particular, I will write 1 for the unit element of each of these three groups. so g f satisfies the unitality axiom. (g f )(1) = g( f (1)) = g(1) = 1,
ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS.3 For any x, y F, we have (g f )(xy) = g( f (xy)) = g( f (x) f (y)) = g( f (x))g( f (y)), so g f satisfies the commuting with the group operation axiom. Proposition 1.4. Let G be a group. Then the identity function id G : G G is a group homomorphism. Proof. Clearly id G sends the unit element to the unit element, and commutes with the group operation on G! Propositions 1.3 and 1.4 allow us to make the following definition: Definition 1.5. Let G, H be groups, and let f : G H and g : H G be group homomorphisms. We say that g is inverse to f if g f = id G and f g = id H. If a group homomorphism f : G H has an inverse, then we say that f is an isomorphism of groups (or sometimes just an isomorphism). If G, H are groups and there exists an isomorphism between them, we say that G and H are isomorphic. Definition 1.5 closely resembles the definition of inverses of functions between sets. You know (from previous lectures and/or lecture notes) that a function between sets has an inverse if and only if it is bijective. It s natural to ask if the same thing is true of groups, and indeed, it is: Proposition 1.6. Let G, H be groups, and let f : G H be a group homomorphism. Then f is an isomorphism if and only if f is bijective. Proof. Suppose that f is an isomorphism. Then f has an inverse group homomorphism f 1, and since f f 1 = id H and f 1 f = id G, f 1 is inverse to f as a function between sets (i.e., forget that G and H are groups, and just remember their underlying sets; then f, f 1 are still inverse to one another). So f is bijective. Conversely: assume that f is a bijective group homomorphism. We want to show that f is an isomorphism. Since f is bijective, as a function between sets f has an inverse function f 1 : H G. We just need to show that f 1 is a group homomorphism. If we use multiplicative notation for G and H, then we have f ( f 1 (1)) = 1 since f f 1 = id H. Now f is a group homomorphism, so f (1) = 1, so we have the equation f ( f 1 (1)) = f (1); and since f is assumed to be bijective, in particular injective, we have f 1 (1) = 1. So f 1 satisfies the unitality axiom. Now let s check the commutes with the group operation axiom. Let x, y H. Then f ( f 1 (x) f 1 (y)) = f ( f 1 (x)) f ( f 1 (y)) = xy, and when we apply f 1 to those equalities, we get f 1 (x) f 1 (y) = f 1 ( f ( f 1 (x) f 1 (y))) = f 1 (xy), so f 1 satisfies the commutes with the group operation axiom. So f 1 is indeed a group homomorphism, so f is a group isomorphism.
4 ANDREW SALCH The idea behind an isomorphism of groups is that you may sometimes find yourself with two groups which aren t literally equal, but there is a one-to-one correspondence between them which preserves the group operation, so they are structured in the same way as groups, but with the elements named differently and that is the intuition behind the idea of two groups being isomorphic. For example: let A be the set of all nonzero rational numbers of the form 2m 2, under n multiplication, and let Z be the group of integers, under addition. An element of A is given by an integer power of 2, while an element of Z is given by an integer, and the function f : Z A defined by f (x) = 2 x is a group homomorphism as well as being a one-to-one correspondence. So the groups A and Z are isomorphic, although they are obviously not equal: the elements of Z are not literally the same as the elements of A. 2. Subgroups and quotient ( factor ) groups. Definition 2.1. Let G be a group. I will use multiplicative notation for G. A subgroup of G is a subset S of G satisfying the properties: the identity element 1 G is contained in S, given two elements x, y S, their product xy in G is also contained in S, and given an element x S, its inverse x 1 in G is also contained in S. In other words: a subgroup of G is a subset of G which contains the unit element and is closed under the group operation and under inverses. Or in other words: a subgroup of G is a subset S of G, equipped with a group structure, and such that the inclusion function of the subset S G into G is a group homomorphism. Example 2.2. The even integers, under addition, form a subgroup of the integers Z under addition. We often write 2Z for the even integers as a subgroup of Z. More generally: given a nonzero integer n, the set of integers divisible by n, under addition, forms a subgroup of Z. We usually write nz for this subgroup. Definition 2.3. Let G be a group (written multiplicatively), let H be a subgroup of G, and let g G be an element. We write g 1 Hg for the subgroup {g 1 hg : h H} of G; this subgroup of G is called the conjugate of H by g. (Checking that g 1 Hg is a subgroup, not just a subset, of G is easy: we have g 1 1g = 1, so the unit element of G is contained in g 1 Hg, and we have (g 1 xg)(g 1 yg) = g 1 (xy)g, so g 1 Hg is closed under multiplication in G.) Definition 2.4. Let G be a group (written multiplicatively) and let H be a subgroup of G. We say that H is a normal subgroup of G if g 1 Hg = H for all g G. Example 2.5. If G is abelian, then every subgroup of G is normal, since g 1 hg = g 1 gh = h. Example 2.6. Recall the symmetric group on 3 letters, written Σ 3 or S 3, from lecture notes #4. Let S be the subset {e, (12)} of Σ 3. Since (12)(12) = e, the subset S is actually not just a subset, but in fact is a subgroup of Σ 3. I claim that S is a non-normal subgroup of Σ 3. Here is a quick computation which shows this: the inverse of (123) in Σ 3 is (321) = (132), so we have equalities (123) 1 (12)(123) = (132)(12)(123) = (23)(123) = (13), and consequently (123) 1 S (123) = {e, (13)} S. So S is not a normal subgroup of Σ 3.
ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS.5 Example 2.7. Let T be the subset {e, (123), (132)} of Σ 3. Since (123) 2 = (132) and (123)(132) = (123) 3 = e, T is in fact a subgroup of Σ 3. I claim that T, unlike S, is a normal subgroup of Σ 3. Here is a quick computational proof: first, since T itself is abelian, every element in T commutes with every element in T, so x 1 T x = T for all x T. So we only need to check the effect of conjugating T by elements of Σ 3 not contained in T. Here they are: (12) 1 (123)(12) = (13)(12) = (132), (12) 1 (132)(12) = (23)(12) = (123), (13) 1 (123)(13) = (23)(13) = (132), (13) 1 (132)(13) = (12)(13) = (123), (23) 1 (123)(23) = (12)(23) = (132), (23) 1 (132)(23) = (13)(23) = (123). So (12) 1 T(12) = T, and (13) 1 T(13) = T, and (23) 1 T(23) = T. So T is a normal subgroup of Σ 3. There are quicker ways to check for normality of a subgroup which we will get to when we cover generators and relations (probably the next lecture). Definition 2.8. Let G be a group, written multiplicatively, and let N be a subgroup of G. By a right coset of N in G we mean an equivalence class of elements in G, under the equivalence relation given by letting x y if and only if there exists some element n N such that xn = y. Given an element x G, we often write xn for the right coset of N in G which contains x. Proposition 2.9. Let G be a group, written multiplicatively, and let N be a normal subgroup of G. Then the set of right cosets of N in G, under the binary operation (xn)(yn) = (xy)n, forms a group, which we will call G/N, the quotient group of G by N (but your textbook uses an old-fashioned terminology, the factor group of G by N). Furthermore, the function G G/N sending x to xn is a surjective group homomorphism. Proof. First, we have to check that the operation (xn)(yn) = (xy)n is well-defined. Suppose we have two elements x 0, x 1 G such that x 0 N = x 1 N, and suppose similarly that we have two elements y 0, y 1 G such that y 0 N = y 1 N. We need to know that (x 0 N)(y 0 N) = (x 1 N)(y 1 N). Since x 0 N = x 1 N, we can choose an element n N such that x 0 n = x 1, and since y 0 N = y 1 N, we can choose an element n N such that y 0 n = y 1.
6 ANDREW SALCH Then we have equalities x 1 y 1 = x 0 ny 0 n = x 0 y 0 (y 1 0 ny 0)n = x 0 y 0 n n for some n N, because N is a normal subgroup of G, so y 1 0 Ny 0 = N (this is why normality of subgroups matters! If we don t assume N is a normal subgroup of G, then we don t get a well-defined group operation on the set of cosets of N in G!!). Now n n N since N is a subgroups and hence closed under the group operation in G. So x 1 y 1 = x 0 ny 0 n x 0 y 0 N, i.e., (x 0 N)(y 0 N) = x 0 y 0 N = x 1 y 1 N = (x 1 N)(y 1 N), as desired. Now we need to check that this binary operation (xn)(yn) = (xy)n on the set of cosets of N in G is unital, associative, and has inverses: Unitality: The coset 1N certainly satisfies (1N)(xN) = (1x)N = xn = (x1)n = (xn)(1n) for all x G, so 1N is a unit element for G/N. Associativity: For any x, y, z G, ((xn)(yn))(zn) = (xyn)(zn) = ((xy)z)n = (x(yz))n = (xn)((yz)n) = (xn)((yn)(zn)), as desired. Inverses: Given x G, the coset x 1 N satisfies (xn)(x 1 N) = xx 1 N = 1N = x 1 xn = (x 1 N)(xN), so x 1 N is an inverse for xn. So G/N is a group. The function f : G G/N given by f (x) = xn certainly sends the unit element to the unit element, so to check that f is a group homomorphism we only need to check that f commutes with the group operation. But this is very easy: f (xy) = (xy)n = (xn)(yn) = f (x) f (y). That f is surjective is similarly easy: given a coset xn in G/N, clearly xn = f (x). Example 2.10. Let n be a nonzero integer and let nz be the subgroup of Z (under addition) consisting of all integers divisible by n. Since Z is abelian, all subgroups of Z are normal, and in particular nz is normal. Two integers a, b are in the same coset of nz if and only if there exists some element z nz such that a+z = b; in other words, two integers a, b are in the same coset of nz if and only if a b is a multiple of n. So..., 2n, n, 0, n, 2n, 3n, 4n,... is one element (in fact the unit element) of Z/nZ, while..., 2n + 1, n + 1, 1, n + 1, 2n + 1, 3n + 1, 4n + 1,... is another element, etc.; and given two cosets a + nz and b + nz, their sum in Z/nZ is a + b + nz. In other words: the quotient group Z/nZ is exactly the group of integers modulo n, under addition modulo n.