Math 3121 Abstract Algebra I Lecture 11 Finish Section 13 Section 14

Next Midterm Midterm 2 is Nov 13. Covers sections: 7-14 (not 12) Review on Thursday

Section 13 Homomorphisms Definition of homomorphism (recall) Examples Properties Kernel and Image Cosets and inverse images Monomorphisms Normal Subgroups

Images and Inverse Images Let X and Y be sets, and let f: X  Y Define f[A] and f-1[B] for subsets A of X and B of Y: f[A] = { b in Y | b = f(a), for some a in A} f-1[B] = { a in X | f(a) is in B}

Properties of Homomorphisms Theorem: Let h be a homomorphism from a group G into a group G’. Then 1) If e is the identity in G, then h(e) is the identity in G’. 2) If a is in G, then h(a-1) = (h(a))-1 3) If H is a subgroup of G, then f[H] is a subgroup of G’. 4) If K’ is a subgroup of G’, then h-1[K’] is a subgroup of G. Proof: Straightforward – in class and in the book

Kernel Definition: Let h be a homomorphism from a group G into a group G’. The kernel of h is the inverse image of the trivial subgroup of G’: Ker(h) = { x in G | h(x) = e’}

Examples of Kernels Modulo n: Z  Zn, x ↦ x + nZ Parity: Sn  Z2 Multiply by m: Zn  Zn, x ↦ mx n = 6, m = 1, 2, 3

Cosets of the kernel are inverse images of elements Theorem: Let h be a homomorphism from a group G into a group G’. Let K be the kernel of h. Then a K = {x in G | h(x) = h(a)} = h -1[{h(a)}] and also K a = {x in G | h(x) = h(a)} = h -1[{h(a)}] Proof: h -1[{h(a)}] = {x in G | h(x) = h(a)} directly from the definition of inverse image. Now we show that: a K = {x in G | h(x) = h(a)} : x in a K ⇔ x = a k, for some k in K ⇔ h(x) = h(a k) = h(a) h(k) = h(a) , for some k in K ⇔ h(x) = h(a) Thus, a K = {x in G | h(x) = h(a)}. Likewise, K a = {x in G | h(x) = h(a)}.

Equivalence Relation Suppose: h: X  Y is any map of sets. Then h defines an equivalence relation ~h on X by: x ~h y ⇔ h(x) = h(y) The previous theorem says that when h is a homomorphism of groups then the cosets (left or right) of the kernel of h are the equivalence classes of this equivalence relation.

Monomorphisms and Epimorphisms Recall: A homomorphism h: G  G’ is called a monomorphism if it is 1-1. A homomorphism h: G  G’ is called an epimorphism if it is onto.

Monomorphism Test Theorem: A homomorphism h is 1-1 if and only if Ker(h) = {e}. Proof: Let h: G  G’ be a homomorphism. Then h(x) = h(a) ⇔ x  a Ker(h). If Ker(h) = {e}, then a Ker(h) = {a} and h(x) = h(a) ⇔ x = a. If Ker(h) is larger, then there is an k different from e in Ker(h), then ak ≠ a and h(ak) = h(a). So h is not 1-1.

Isomorphism Test To show h : G  G’ is an isomorphism Show h is a homomorphism Show Ker(h) = {e} Show h is onto.

Normal Subgroups Definition: A subgroup H of a group G is said to be normal if a H = H a, for all a in G.

Kernel is Normal Theorem: Let h: G  G’ be a group homomorphism, then Ker(h) is normal: Proof: By previous theorem, a Ker(h) = Ker(h) a, for all a in G. By the previous definition, Ker(h) is normal.

HW Not to hand in: Hand in (due Thurs Nov 18) Page 133: 1, 3, 5, 7, 17, 19, 27, 29, 33, 35 Hand in (due Thurs Nov 18) Page 133: 44, 45, 49

Section 15 Section 15: Factor Groups Multiplication of cosets Definition: Factor Group Theorem: The image of a group homomorphism is isomorphic to the group modulo its kernel. Properties of normal subgroups Theorem: For a subgroup of a group, left coset multiplication is well-defined if and only if the subgroup is normal. Theorem: The canonical map is a homomorphism.

Multiplication of Cosets Let H be a subgroup of a group G. When is (a H) (b H) = a b H? This is true for abelian groups, but not always when G is nonabelian. Consider S3: Let H = {ρ0, μ1}. The left cosets are {ρ0, μ1}, {ρ1, μ3}, {ρ2, μ2}. If we multiply the first two together, then {ρ0, μ1}, {ρ1, μ3} = {ρ0 ρ1, ρ0 μ3, μ 1 ρ1, μ 1 μ3} = {ρ1, μ3, μ2, ρ 2} This has four distinct elements, not two!

Sometimes it does work. Consider S3: Let H = {ρ0, ρ1 , ρ2}. The left cosets are {ρ0, ρ1 , ρ2}, {μ1, μ2 , μ3} If we multiply the first two together, then {ρ0, ρ1 , ρ2} {μ1, μ2, μ3} = {ρ0 μ1, ρ0 μ2, ρ0 μ3, ρ1 μ1, ρ1 μ2, ρ1 μ3, ρ2 μ1, ρ2 μ2, ρ2 μ3} = {μ1, μ2, μ3, μ3, μ1, μ2, μ2, μ3, μ1} = {μ1, μ2, μ3} This is one of the cosets. Likewise, {ρ0, ρ1 , ρ2} {ρ0, ρ1 , ρ2} = {ρ0, ρ1 , ρ2} {μ1, μ2 , μ3}{ρ0, ρ1 , ρ2} = {μ1, μ2 , μ3} {μ1, μ2 , μ3 }{μ1, μ2 , μ3} = {ρ0, ρ1 , ρ2} Note that the cosets of {ρ0, ρ1 , ρ2} with this binary operation form a group isomorphic to ℤ2.

Canonical Homomorphism Note that there is a natural map from S3 from {{ρ0, ρ1 , ρ2}, {μ1, μ2 , μ3}} that takes any element to the coset that contains it. This gives a homomorphism called the cannonical homomorphism.

Theorem Theorem: Let h: G  G’ be a group homomorphism with kernel K. Then the cosets of K form a group with binary operation given by (a K)(b K) = (a b) K. This group is called the factor group G/K. Additionally, the map μ that takes any element x of G to is coset xH is a homomorphism. This is called the canonical homomorphism. Proof: Let (a K)(b K) = { a k1 b k2 | k1,k2 in K}. We show this is equal to (a b) K. Clearly, a b K  (a K)(b K) (just consider what happens when k1 = e) To prove the reverse apply h: h[(a K)(b K)] = { h(a k1 b k2 )| k1,k2 in K} But h(a k1 b k2)= h(a) h( k1) h(b) h(k2 ) = h(a) e’ h(b) e’= h(a) h(b) = h(a b) Then h[(a K)(b K)] = {h(a b)| k1,k2 in K}= {h(a b)} Thus (a K)(b K)  h-1[{h(a b)}] = a b K So (a K)(b K) = a b K.

Associativity of Coset Multiplication Proof continued: This operation is associative: ((a K) (b K)) (c K) = (a b K) (c K) = a b c K (a K)((b K) (c K)) = (a K) (b c K) = a b c K Thus ((a K) (b K)) (c K) = (a K)((b K) (c K))

Identity and Inverse Proof continued: The coset e K = K is an identity: (e K) (a K) = (e a) K = a K For each coset a K, the coset a-1 K is an inverse: (a-1 K) (a K) = (a-1 a) K = e K (a K) (a-1 K) = (a a-1) K = e K

Canonical Map Proof continued: Let μ(a) = a K. Then μ(a b) = a b K and μ(a) μ (b) = (a K)(b K) = a b K Thus μ(a b) = μ(a) μ (b)

Terminology Let H be a subroup of a group G. When the cosets satisfy the rule (a H) (b H) = ( a b) H We call the set of cosets the factor group and denote it by G/H. This is read G modulo H. Note that for finite groups order(G/H) = order(G)/order(H)

Coset Multiplication is equivalent to Normality Theorem: Let H be a subgroup of a group G. Then H is normal if and only if (a H )( b H) = (a b) H, for all a, b in G Proof: Suppose (a H )( b H) = (a b) H, for all a, b in G. We show that a H = H a, for all a in H. We do this by showing: a H  H a and H a  a H, for all a in G. a H  H a: First observe that a H a-1  (a H )( a-1 H) = (a a-1) H = H. Let x be in a H. Then x = a h, for some h in H. Then x a-1 = a h a-1, which is in = a H a-1 , thus in H. Thus x a-1 is in H. Thus x is in H a. H a  a H: H a  H a H = (e H )( a H) = (e a) H = a H. This establishes normality. For the converse, assume H is normal. (a H )( b H)  (a b) H: For a, b in G, x in (a H )( b H) implies that x = a h1 b h2, for some h1 and h2 in H. But h1 b is in H b, thus in b H. Thus h1 b = b h3 for some h3 in H. Thus x = a b h3 h2 is in a b H. (a b) H  (a H )( b H): x in (a b) H ⇒that x = a e b h, for some h in H. Thus x is in (a H) (b H).