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WEEK 6: MAPS on GROUPS A. Homomorphisms A.1. Definitions. Naturally We Can Consider Functions Mapping One Group to Another in Ex

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WEEK 6: MAPS on GROUPS A. Homomorphisms A.1. Definitions. Naturally We Can Consider Functions Mapping One Group to Another in Ex WEEK 6: MAPS ON GROUPS A. Homomorphisms A.1. Definitions. Naturally we can consider functions mapping one group to another in exactly the same way that we may look at functions relating any two sets. But if we gain nothing by limiting our discussion to groups unless the functions interact with the group structure in some way. Let G and H be groups. A function f : G ! H is a homomorphism if f(ab) = f(a)f(b): Note that the product on the left is with the operation on G, while that on the right is the operation on H. We might say that the map f preserves the operation. These maps need not be injective or surjective. For example, consider the following map from Z to D3: 8 r ; if n = 3q; <> 0 f(n) = r1; if n = 3q + 1; and > :r2; if n = 3q + 2; where q is some integer. Let's check to see that f : Z ! D3 is a homomorphism. If n = 3q + j and m = 3s + t, with 0 ≤ j; t < 3, we have f(n + m) = f ((3q + j) + (3s + t)) = f (3(q + s) + (j + t)) : The image of this element depends entirely on j +t: if j +t is divisible by 3, then f(n+m) = r0, if j + t − 1 is divisible by 3, then f(n + m) = r1, and if j + t − 2 is divisible by 3, then f(n + m) = r2. Thus, f(n+m) = ri if j +t = i mod 3. But the operation on D3 is such that ri = rjrt if j +t = i mod 3. So f(n + m) = rjrt = f(n)f(m), and f is a homomorphism. This particular example is neither injective (f(0) = f(3)) nor is it surjective (nothing maps to s0; s1; or s2). Exercise A.1.1. Show that fd : Z ! Z, given by fd(n) = dn is, for any fixed integer d, a homomorphism. Then show that it is injective. For what values of d is it surjective? Lemma A.1.2. Let eG and eH denote the identities in the groups G and H, respectively. If f : G ! H is a homomorphism, then f(eG) = eH . Exercise A.1.3. Prove Lemma A.1.2. Corollary A.1.4. Let f : G ! H be a homomorphism. If f(g) = h, then f(g−1) = h−1. Proof. From Lemma A.1.2, we know that −1 −1 f(g)f(g ) = f(gg ) = f(eG) = eH : 1 Since f(g) = h and from the equation above we have that −1 hf(g ) = eH ; −1 −1 f(g ) must be h . + Discussion Question A.1.5. Let R denote the positive reals. The same function can be a homomorphism under one operation and not a homomorphism under another. The opera- tion makes the group{ the same elements under different operations can form groups that are completely different. • Show that f : R ! R, f(x) = 2x is a homomorphism as long as R is considered as a group under addition, + + + • but g : R ! R , g(x) = 2x is not a homomorphism when we consider R as a group under multiplication. Discussion Question A.1.6. If a group is finitely generated, then tracking where the genera- tors go under a given map is of particular interest. For example, where did 1 map to in Exercise A.1.1? Exercise A.1.7. Let hai be the subgroup generated by a single element of the symmetric generating set of the free group on two elements, F2. Show that there is a homomorphism mapping hai to Z (under addition). A.2. Kernel and image. The kernel of a homomorphism is the set of all elements that it maps to the identity. That is, if f : G ! H is a homomorphism, and eH is the identity in H, then Ker(f) = fg 2 G : f(g) = eH g: The image of a set A ⊆ G under the homomorphism f : G ! H is the set of all elements in H that elements from A map to. In set notation, f(A) = ff(a) 2 H : a 2 Ag: We use the phrase image of f to denote the image of G under f. Thus Im(f) = f(G) = ff(g) 2 H : g 2 Gg: f is surjective, or onto, if Im(f) = H. You can think of the image of f as a subset of the codomain, identical to what we usually would call the range. Exercise A.2.1. Show that if f : G ! H is a homomorphism then Im(f) is a subgroup of H. Theorem A.2.2. Let f : G ! H be a homomorphism of groups. f is injective (one-to-one) if, and only if, Ker(f) = feH g. Proof. If f is injective, then it is clear that Ker(f) = feGg. −1 −1 Suppose that Ker(f) = feH g and f(g1) = f(g2) = h. Then (by Corollary A.1.4) h = f(g2 ) and −1 f(g1)f(g2 ) = eH : Because f is a homomorphism we have that −1 f(g1g2 ) = eH : −1 This means that g1g2 2Ker(f). −1 But the kernel of f contains only the identity on G, so g1g2 = eG. 2 Applying g2 to the right of both sides of this equation, we find that g1 = g2: So f is injective. Discussion Question A.2.3. We committed the venial sin, in the first line, of saying that something is \clear." Why is it \clear" that if f is injective then Ker(f) = feGg? Discussion Question A.2.4. It is worth some time to examine, again, how one can show injectivity. Our strategy here is to show that if f(a) = f(b) then a must equal b. Why is this enough? B. Isomorphisms B.1. Definitions and properties. A bijective (one-to-one and onto) homomorphism is called an isomorphism. An isomorphism on groups is a renaming of each element of the group. As far as the group itself is concerned, it does not matter if what you call each element{ how they interact is what matters. We saw this when we adopted different notations for elements of S3. The objects 1 2 3 a b c and 1 3 2 a c b are exactly the same element in S3. The isomorphism f : G ! H takes elements of G and gives them new unique (thanks to injectivity) names. Each g 2 G becomes f(g). By virtue of the fact that f is a homomorphism, these elements interact in exactly the same way as they did before, just with new names. And because every available name is used (surjectivity), we find that H, as a group, is the same thing as G. If there is an isomorphism from G to H (f : G ! H) we say that H and G are isomorphic and write G =∼ H: This is a reflexive, symmetric, and transitive relation. It is thus an equivalence relation. Discussion Question B.1.1. To show that this is a reflexive relation, show that the identity map on G is an isomorphism. Theorem B.1.2. If f : G ! H is an isomorphism, then f −1 is an isomorphism mapping H to G. Proof. Since f is an injective function, f −1 is also a function and injective. Since f is surjective, f −1 takes all of H as its domain and, because it is an inverse, is surjective, mapping H onto G. All that remains is to show that f −1 is a homomorphism. Let f(a) = h and f(b) = k. Then f −1(h) = a and f−1(k) = b. We are done if f −1(hk) = ab = f −1(h)f −1(k): Since f(ab) = f(a)f(b) = hk, we apply the inverse function to both sides of the equation to complete the proof. Theorem B.1.3. If G =∼ H: and H =∼ K, then G =∼ K. Exercise B.1.4. Prove Theorem B.1.3. You'll need to show that • the composition of injective functions is injective, 3 • the composition of surjective functions is surjective, and • the composition of homomorphisms is a homomorphism. B.2. Examples. + Exercise B.2.1. Let h2i be the subgroup generated by 2 of Q { the positive rationals under multiplication. Show that this group is isomorphic to Z under addition. Discussion Question B.2.2. Is every infinite group generated by a single element isomorphic to Z under addition? Exercise B.2.3. Show that Z3 under addition is isomorphic to hri ⊆ D3. Exercise B.2.4. Show that Z3 under addition is isomorphic to 2πi he 3 i ⊆ C (under multiplication). Discussion Question B.2.5. Is every cyclic group of order 3 isomorphic to Z3 under addition? Exercise B.2.6. The set 1 0 1 1 0 1 0 1 1 1 1 0 GL ( ) = ; ; ; ; ; 2 Z2 0 1 0 1 1 0 1 1 1 0 1 1 forms a group under matrix multiplication (see Discussion B.1.5, Week 2), with addition modulo 2. For example, 1 1 1 1 1 + 0 1 + 1 1 0 = = : 0 1 0 1 0 + 0 0 + 1 0 1 Show that this group is isomorphic to S3. One approach would be to map 1 1 0 1 and 0 1 1 0 to transpositions{ say, t1;2 and t1;5{ then use the fact that these two transpositions generate S3 to determine where the other matrices map. Confirm that the map is a homomorphism using multiplication tables for each group.
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