Lecture 5: The Correspondence Theorem

Lecture 5: The Correspondence Theorem 5 The Correspondence Theorem 5.1 Review In the last lecture, we learned about cosets and some of their properties. Defnition 5.1 For a group G and a subgroup H ≤ G, we defne the left coset of a to be aH := {ah : h ∈ H} ⊆ G. The left cosets partition 22 G into equally sized sets. This provides a useful corollary about the structure of cosets within a group: Corollary 5.2 (Counting Formula.) Let [G : H] be the number of left cosets of H, which is called the index of H in G. Then |G| = |H|[G : H]. 5.2 Lagrange’s Theorem Using cosets provides some additional information about groups. Guiding Question What are the possibilities for the structure of a group with order n? From the Counting Formula, we immediately obtain Lagrange’s Theorem as a corollary: Theorem 5.3 (Lagrange’s Theorem.) For H a subgroup of G, |H| is a divisor of |G|. Several important corollaries follow as a result. Corollary 5.4 The order of x ∈ G is |⟨x⟩|. Since the order of any subgroup divides the order of |G|, ord(x) also divides |G|. Corollary 5.5 Any group |G| with prime order p is a cyclic group. Proof. Take an element e ̸= x ∈ G. Since the order of x ∈ G divides p, and p is prime, ord(x) = p. Then each xi is distinct for 0 ≤ i ≤ p − 1, and since there are only p elements in G, the entire group G is ⟨x⟩, the cyclic group generated by x. Our result shows that any group of prime order is a cyclic group. In particular, the integers modulo p, Zp , form a cyclic group of prime order; that is, any group of prime order p is isomorphic to Zp . 5.3 Results of the Counting Formula Using Lagrange’s Theorem narrows down the possibilities for subgroups. 22 A partition of a set S is a subdivision of the entire set into disjoint subsets. Lecture 5: The Correspondence Theorem Example 5.6 (Groups of order 4.) What are the possibilities (up to isomorphism) for G if |G| = 4? First, e must be an element of G. Next, consider the other three elements of G. Each of these must have either order 2 or order 4, since those are the divisors of |G| = 4. Then there are two possibilities. • Case 1. There exists an element x ∈ G such that ord(x) = 4. Then we know that e ̸= x ̸= x2 = ̸ x3 , 4 and since |G| = 4, these are all the elements of G. (The power x is e again.) So G is generated by x, and it is the cyclic group ⟨x⟩ of size 4, and must be isomorphic to Z4 a . • Case 2. All elements of G have order 2. Then, we can take x ∈ G and y = ̸ x ∈ G. They have order 2, so x2 = e, which implies that x = x−1 and similarly y = y −1 . Also, the element xy also has order 2, and so xyx−1 y −1 = (xy)(xy) = e, and so x and y commute. Because x and y were chosen arbitrarily, any two elements of the group commute, and so it is abelian. This group G is isomorphic to the matrix group   ±1 0 ≤ GL2 (R). 0 ±1 The non-identity elements each have order 2 and commute with each other. This group is called the Klein-four group, and is denoted K4 . Up to isomorphism, any order 4 group is either Z2 or K4 . Note that both of these groups are abelianb ; the smallest non-abelian group has order 6. a We write Zn to b commutative denote the group of integers modulo n. Exercise 5.7 What are the possible groups of order 6? The Counting Formula also provides another important corollary. Corollary 5.8 The size of the group is |G| = |ker(f )| · |im(f )|.a a In linear algebra, the analogous result is the rank-nullity theorem. Proof. Let f : G → G′ be a homomorphism, and ker(f ) ≤ G be the kernel. For each y ∈ G′ , the preimage of y is f −1 (y) := {x ∈ G : f (x) = y}, which is ∅ if y ∈ / im(f ), and a coset of ker(f ) otherwise.23 Then, the number of left cosets of ker(f ) is precisely the number of elements in the image of f, since each of those elements corresponds to a coset of the kernel. So [G : ker(f )] = |im(f )|, and applying the counting formula with ker(f ) as our subgroup H gives us |G| = |ker(f )| · |im(f )|, which is the desired result. 5.4 Normal Subgroups In this section, we learn about normal subgroups. 23 Pick x ∈ f −1 (y). Then we claim that f −1 (y) = x ker(f ). Take any x′ ∈ f −1 (y). We have y = f (x) = f (x′ ) = f (xx′−1 )f (x′ ), so f (xx′−1 ) = e and xx′−1 ∈ ker(f ). Thus x′ ∈ x ker(f ). Lecture 5: The Correspondence Theorem Guiding Question The choice of left cosets seems arbitrary — what are the ramifcations if right cosets are used instead? Defnition 5.9 The right coset of a is Ha = {ha : h ∈ H}. In fact, all the same results follow if right cosets are used instead of left cosets. First, let’s see an example of right cosets: Example 5.10 Let H be the subgroup generated by y ∈ S3 . Then the left cosets are {e, y}, {x, xy}, {x2 , x2 y}, and the right cosets are {e, y}, {x, x2 y}, {x2 , xy}. So in fact, right cosets give a diferent partition of S3 , but the number and size of the cosets are the same.a a We can think of cosets as "carving up" the group. Using right cosets instead of left cosets is just carving it up in a diferent way. In particular, there is a bijection between the set of left cosets and the set of right cosets. It maps C 7→ C −1 = {x−1 : x ∈ C}. It is a bijection because (ah)−1 = h−1 a−1 , and so aH = Ha−1 . So the index [G : H] is equal to both the number of right cosets and the number of left cosets. Guiding Question For which subsets H ⊆ G do left and right cosets give the same partition of G? In other words, for which H is every left coset also a right coset?a a If some left coset xH of an element x is equal to some right coset Hy of a diferent element y, since x ∈ Hy as well, from a lemma from last week’s lecture, Hy = Hx, and so in fact the left coset and right coset of the same element x must also be equal. So it is sufcient to require that xH = Hx. This question motivates the defnition of normal subgroups. Defnition 5.11 If xH = Hx for each x ∈ G, H ⊆ G is called a normal subgroup. Equivalently, the subgroup H is normal if and only if it is invariant under conjugation by x; that is, xHx−1 = H. Using the notation from last lecturea , a subgroup H is normal if and only if φx (H) = H for all x ∈ G. a The function φx takes g 7→ xgx−1 . Let’s look at some examples. Example 5.12 (Non-normal subgroup) From above, the subgroup ⟨y⟩ is not normal in S3 . Lecture 5: The Correspondence Theorem Example 5.13 (Kernel) Given a homomorphism f : G → G′ , the kernel of f is always normal. Take k ∈ ker(f ). Then f (xkx−1 ) = f (x)f (k)f (x)−1 = f (k) = eG′ , so φx (ker(f )) = ker(f ), and thus ker(f ) is a normal subgroup. In fact, in future lectures, we will see that all normal subgroups of a given group G arise as the kernel of some homomorphism f : G → G′ to a group G′ . Example 5.14 In S3 , the subgroup ⟨y⟩ is not normal, but ⟨x⟩ is normal. In particular, it is the kernel of the sign homomorphism sign : S3 → R.a a A given permutation σ can be written as a product of i transpositions, where i is unique up to parity. The sign homomorphism maps σ to (−1)i . 5.5 The Correspondence Theorem Ealier in this lecture, we noticed that homomorphisms give us some information about subgroups. Can we make this more concrete? Guiding Question Let f be a homomorphism from G to G′ . Is there a relationship between the subgroups of G and the subgroups of G′ ? {subgroups of G} ↔ {subgroups of G′ } Answer. In fact, we see that there is! • Given a subgroup of G, a subgroup of G′ can be produced as follows. Let f with the domain restricted to H be denoted as f |H . Then a subgroup H ≤ G maps to im(f |H ) = f (H) ⊆ G′ , which is a subgroup of G′ . • Now, given H ′ ≤ G′ and a subgroup of G can be produced by taking the preimage f −1 (H ′ ) = {x ∈ G : f (x) ∈ H ′ }. Is this subset of G is actually a subgroup? It is! Let’s just check that it’s closed under composition. If x, y ∈ f −1 (H), then f (x), f (y) ∈ H ′ , so f (x)f (y) ∈ H ′ , since H ′ is closed under multiplication. Then f (xy) ∈ H ′ , so xy ∈ f −1 (H). If H ′ = eG′ , then its preimage is the kernel, and if H ′ = G′ , then the preimage is all of G. In general, the preimage is a subgroup somewhere in-between the kernel and the whole domain. Are these maps bijective, or inverses of each other? It can be easily seen that they are not; in particular, if G is the trivial group and G′ is some more complicated group with many subgroups, every subgroup of G′ must always still map to the trivial group. It makes sense that these maps are not bijective, since f is not an isomorphism, just an arbitrary homomorphism with no more restrictions. Two issues arise with these maps that make them non-bijective: • Any subgroup of G must map to some subgroup of G′ that is contained within the image of f, by construction, since f (H) ⊆ im(f ). • The kernel ker(f ) = f −1 (eG′ ) ⊆ f −1 (H ′ ), so any subgroup not contained within the kernel cannot be mapped to by any subgroup of G′ . However, these are actually the only issues! If we are willing to put some restrictions on the homomorphism f and the types of subgroups we look at, there is actually a bijection between certain subgroups of G and certain subgroups of G′ . Lecture 5: The Correspondence Theorem In order to make things a little easier for now, we take a surjective homomorphism f : G → G′ . The frst issue then is no longer consequential, because the image is all of G′ . Now, let’s restrict the subgroups of G to subgroups that contain ker(f ). Then our maps (as described above) provide a bijection. Theorem 5.15 (Correspondence Theorem) For a surjective homomorphism f with kernel K, there is a bijective correspondence: {subgroups of G containing K} ↔ {subgroups of G′ }, where a subset of G, H ⊇ K ⇝ its image f (H) ≤ G′ H ′ ≤ G′ ⇝ its preimage f −1 (H ′ ) ≤ G. Example 5.16 (Roots of Unity) Take f G = C∗ − → G′ = C∗ z 7→ z 2 , which is a homomorphism because G is abelian. The kernel is ker(f ) = {±1}. We have a correspondence between R × ⇝ R>0 . For example, the eighth roots of unity correspond to the fourth roots of unity under this map. H = {e 2πik 8 } ↭ H ′ = {±1, ±i}.