Lecture 4: Isomorphisms and Cosets

Lecture 4: Isomorphisms and Cosets 4 Isomorphisms and Cosets 4.1 Review In the last lecture, we learned about subgroups and homomorphisms. Defnition 4.1 We call f : G → G′ a homomorphism if for all a, b ∈ G, f (a)f (b) = f (ab). Defnition 4.2 The kernel of a homomorphism f is {a ∈ G : f (a) = eG′ }, and the image is the set of elements b = f (a) for some a. The kernel and image of f are subgroups of G and G′ , respectively. 4.2 Isomorphisms Homomorphisms are mappings between groups; now, we consider homomorphisms with additional constraints. Guiding Question What information can we learn about groups using mappings between them? Defnition 4.3 We call f : G → G′ an isomorphism if f is a bijective homomorphism. In some sense, if there exists an isomorphism between two groups, they are the same group; relabeling the elements of a group using an isomorphism and using the new product law yields the same products as before relabeling. Almost all the time, it is only necessary to consider groups up to isomorphism. Example 4.4 There exists an isomorphism f : Z4 → ⟨i⟩ given by n mod 4 7→ in . In particular, we get 0 7→ 1 1 7→ i 2 7→ −1 3 7→ −i. So the group generated by i, which can be thought of as a rotation of the complex plane by π/2, is essentially "the same" as the integers modulo 4. Example 4.5 More generally, the group generated by g, ⟨g⟩ = {e, g, g 2 , · · · , g d−1 }, where d is the order of g, is isomorphic to Zd = {0, 1, · · · , d − 1}. If the order of g is infnite, then we have ⟨g⟩ ∼ = Z. Here, the idea that an isomorphism is a "relabeling" of elements makes sense: since g a g b = g a+b , relabeling g i with its exponent i retains the important information in this situation. Thinking of ⟨g⟩ in this way yields precisely Zd . 4.3 Automorphisms An important notion is that of an automorphism, which is an isomorphism with more structure. Lecture 4: Isomorphisms and Cosets Defnition 4.6 An isomorphism from G to G is called an automorphism. If a homomorphism can be thought of as giving us some sort of "equivalence" between two groups, why do we care about automorphisms? We already have an equivalence between G and itself, namely the identity. The answer is that while the identity map id : G → G is always an automorphism, more interesting ones exist as well! We can understand more about the symmetry and structure of a group using these automorphisms. Example 4.7 A non-trivial automorphism from Z to itself is f : Z → Z taking n 7→ −n. From the existence of this nontrivial automorphism, we see that Z has a sort of "refective" symmetry.17 Example 4.8 (Inverse transpose) Another non-trivial automorphism, on the set of invertible matrices, is the inverse transpose f : GLn (R) → GLn (R) A 7→ (At )−1 Many other automorphisms exist for GLn (R),18 since it is a group with lots of structure and symmetry. Example 4.9 (Conjugation) A very important automorphism is conjugation by a fxed element a ∈ G. We let ϕa : G → G be such that ϕa (x) = axa−1 . We can check the conditions to show that conjugation by a is an automorphism: • Homomorphism. ϕa (x)ϕa (y) = axa−1 aya−1 = axya−1 = ϕa (xy). • Bijection. We have an inverse function ϕa−1 : ϕa−1 ◦ ϕa = ϕa ◦ ϕa−1 = id. Note that if G is abelian, then ϕa = id. Any automorphism that can be obtained by conjugation is called an inner automorphism; any group intrinsically has inner automorphisms coming from conjugation by each of the elements (we can always fnd these automorphisms to work with). Some groups also have outer automorphisms, which are what we call any automorphisms that are not inner. For example, on the integers, the only inner automorphism is the identity function, since they are abelian.19 4.4 Cosets Throughout this section, we use the notation K := ker(f ). Guiding Question When do two elements of G get mapped to the same element of G′ ? When does f (a) = f (b) ∈ G′ ? Given a subgroup of G, we can fnd "copies" of the subgroup inside G. 17 In particular, this automorphism f corresponds to refection of the number line across 0. example, just the transpose or just the inverse are automorphisms, and in fact they are commuting automorphisms, since the transpose and inverse can be taken in either order. 19 For an abelian group, axa−1 = aa−1 x = x. 18 For Lecture 4: Isomorphisms and Cosets Defnition 4.10 Given H ⊆ G a subgroup, a left coset of H is a subset of the form aH := {ax : x ∈ H} for some a ∈ G. Let’s start with a couple of examples. Example 4.11 (Cosets in S3 ) Let’s use our favorite non-abelian group, G = S3 = ⟨(123), (12)⟩ = ⟨x, y⟩, and let our subgroup H be {e, y}. Then eH = H = {e, y} = yH; xH = {x, xy} = xyH; and x2 H = {x2 , x2 y} = x2 yH. We have three diferent cosets, since we can get each coset one of two ways. Example 4.12 If we let G = Z and H = 2Z, we get 0 + H = 2Z = evens = 2 + H = · · · , and 1 + H = 1 + 2Z = odd integers = 3 + H = · · · . In this example, the odd integers are like a "copy" of the even integers, shifted over by 1. From these examples, we notice a couple of properties about cosets of a given subgroup. Proposition 4.13 All cosets of H have the same order as H. Proof. We can prove this by taking the function fa : H → aH which maps h 7→ ah. This is a bijection because it is invertible; the inverse is fa−1 .20 Proposition 4.14 Cosets of H form a partition of the group G.a aA partition of a set S is a subdivision of S into disjoint subsets. To prove this, we use the following lemma. Lemma 4.15 Given a coset C ⊂ G of H, take b ∈ C. Then, C = bH. Proof. If C is a coset, then C = aH for some a ∈ G. If b ∈ C, then b = ah for some h ∈ H, and a = bh−1 . Then bH = {bh′ : h′ ∈ H} = {ahh′ |h′ ∈ H} ⊆ aH. Using a = bh−1 , we can similarly show that aH ⊆ bH, and so aH = bH.21 20 I can undo any f in a unique way by multiplying again on the left by a−1 . This is something that breaks down with monoids a or semigroups or other more complicated structures. 21 So for a given coset C, we can use any of the elements in it as the representative a such that C = aH. Lecture 4: Isomorphisms and Cosets Proof. Now, we prove our proposition. • Every x ∈ G is in some coset. Take C = xH. Then x ∈ C. • Cosets are disjoint. If not, let C, C ′ be distinct cosets, and take y in their intersection. Then yH = C and yH = C ′ by Lemma 4.15, and so C = C ′ . With this conception of cosets, we have the answer to our question: Answer. If f (a) = f (b), then f (a)−1 f (b) = eG′ . In particular, f (a−1 b) = eG′ , so a−1 b ∈ K, the kernel of f. Then, we have that b ∈ aK, or b = ak where f (k) = eG′ . So f (a) = f (b) if a is in the same left coset of the kernel as b. 4.5 Lagrange’s Theorem In fact, thinking about cosets gives us quite a restrictive result on subgroups, known as Lagrange’s Theorem. Guiding Question What information do we automatically have about subgroups of a given group? Defnition 4.16 The index of H ⊆ G is [G : H], the number of left cosets. Theorem 4.17 We have |G| = [G : H]|H|. Proof. This is true because each of the cosets have the same number of elements and partition G. So we have |G| = X |C| = left cosets C X |H| = [G : H]|H|. left cosets C That is, the order of G is the number of left cosets multiplied by the number of elements in each one (which is just |H|). Example 4.18 For S3 , we have 6 = 3 · 2. From our theorem, we get Lagrange’s Theorem: Corollary 4.19 (Lagrange’s Theorem.) For H a subgroup of G, |H| is a divisor of |G|. We have an important corollary about the structure of cyclic groups. Corollary 4.20 If |G| is a prime p, then G is a cyclic group. Proof. Pick x ̸= e ∈ G. Then ⟨x⟩ ⊆ G. Since the order of x cannot be 1, since it is not the identity, the order of x has to be p, since p is prime. Therefore, ⟨x⟩ = G, and so G is cyclic, generated by x. In general, for x ∈ G, the order of x is the size of ⟨x⟩, which divides G. So the order of any element divides the size of the group.