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If φ : G Ḡ is a homomorph, g G, and H G: Properties of elements Properties of subgroups 1. φ(e G ) = eḡ 1. φ(h) Ḡ. 2. φ(g n ) = (φ(g)) n for all n Z. 2. H cyclic φ(h) cyclic. 3. If g is finite, φ(g) divides

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If φ : G Ḡ is a homomorph, g G, and H G: Properties of elements Properties of subgroups 1. φ(e G ) = eḡ 1. φ(h) Ḡ. 2. φ(g n ) = (φ(g)) n for all n Z. 2. H cyclic φ(h) cyclic. 3. If g is finite, φ(g) divides g. 3. H Abelian φ(h) Abelian. 4. Ker(φ) G 4. H G φ(h) φ(g) In fact, Ker(φ) G 5. φ(a) = φ(b) aker(φ) = bker(φ) 5. Ker(φ) = n φ is an n-to-1 map 6. φ(g) = ḡ φ 1 (ḡ) = gker(φ) 6. H = n φ(h) divides n 7. K Ḡ φ 1 ( K) G. 8. K Ḡ φ 1 ( K) G. Ker(φ) = {e G } φ is φ onto and Ker(φ) = {e G } φ an isomorphism. 1st Isomorphism Thm: Let φ : G Ḡ be a group homomorphism. Then the mapping φ : G/Ker(φ) φ(g), defined by φ(gker(φ)) = φ(g), is an isomorphism. In other words, G/Ker(φ) φ(g). Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 In Class Work 1. Prove that (A B)/(A {e}) is isomorphic to B, by identifying a homomorphism from A B B that has A {e} as its kernel. 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to Suppose that there is a homomorphism φ from Z 17 to some group, and that φ is not one-to-one. Determine φ. 4. If φ is a homomorphism from Z 30 onto a group of order 5, determine the kernel of φ. 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 1. Prove that (A B)/(A {e}) is isomorphic to B. Define φ : A B B by φ((a, b)) = b. From previous in-class work, we already know that φ is a well-defined onto homomorphism. In fact, we know (more or less) from Friday before break what the kernel is. Ker(φ) = A {e}. The easiest way to do this problem is to use the 1st isomorphism theorem. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 1. Prove that (A B)/(A {e}) is isomorphic to B. Define φ : A B B by φ((a, b)) = b. From previous in-class work, we already know that φ is a well-defined onto homomorphism. In fact, we know (more or less) from Friday before break what the kernel is. Ker(φ) = A {e}. The easiest way to do this problem is to use the 1st isomorphism theorem. The 1st Isomorphism Thm tells us A B/Ker(φ) φ(a B), or A B/Ker(φ) B, since φ is onto. Thus, since Ker(φ) = A {e}, A B/A {e} B. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to 9. We need to find φ 1 (9). Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to 9. We need to find φ 1 (9). Thm 10.1 Property 6 states: if φ(g) = ḡ, then φ 1 (ḡ) = gker(φ). Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to 9. We need to find φ 1 (9). Thm 10.1 Property 6 states: if φ(g) = ḡ, then φ 1 (ḡ) = gker(φ). In Z30, the operation is addition mod 30. Thm 10.1 Prop 5 thus becomes if φ(g) = ḡ, then φ 1 (ḡ) = g + Ker(φ), mod 30. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to 9. We need to find φ 1 (9). Thm 10.1 Property 6 states: if φ(g) = ḡ, then φ 1 (ḡ) = gker(φ). In Z30, the operation is addition mod 30. Thm 10.1 Prop 5 thus becomes if φ(g) = ḡ, then φ 1 (ḡ) = g + Ker(φ), mod 30. We know that φ(23) = 9 Thus φ 1 (9) = 23 + Ker(φ) = 23 + {0, 10, 20} mod 30 Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to 9. We need to find φ 1 (9). Thm 10.1 Property 6 states: if φ(g) = ḡ, then φ 1 (ḡ) = gker(φ). In Z30, the operation is addition mod 30. Thm 10.1 Prop 5 thus becomes if φ(g) = ḡ, then φ 1 (ḡ) = g + Ker(φ), mod 30. We know that φ(23) = 9 Thus φ 1 (9) = 23 + Ker(φ) = 23 + {0, 10, 20} mod 30 = {23, 3, 13}, Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 Solutions: 2. Suppose that φ is a homomorphism from Z 30 to Z 30 and that Ker(φ) = {0, 10, 20}. If φ(23) = 9, determine all elements that map to 9. We need to find φ 1 (9). Thm 10.1 Property 6 states: if φ(g) = ḡ, then φ 1 (ḡ) = gker(φ). In Z30, the operation is addition mod 30. Thm 10.1 Prop 5 thus becomes if φ(g) = ḡ, then φ 1 (ḡ) = g + Ker(φ), mod 30. We know that φ(23) = 9 Thus φ 1 (9) = 23 + Ker(φ) = 23 + {0, 10, 20} mod 30 = {23, 3, 13}, so the elements that map to 9 are 3, 13, and 23. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 3. Suppose that there is a homomorphism φ from Z 17 to some group, and that φ is not one-to-one. Determine φ. Ker(φ) = {e} if and only if φ is 1-1 Thus since φ is not 1-1, Ker(φ) {e}. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 3. Suppose that there is a homomorphism φ from Z 17 to some group, and that φ is not one-to-one. Determine φ. Ker(φ) = {e} if and only if φ is 1-1 Thus since φ is not 1-1, Ker(φ) {e}. Ker(φ) is a subgroup of Z17 which is cyclic of prime order. Since the only subgroups of Z 17 are the trivial subgroup or the whole group, Ker(φ) = Z 17. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 3. Suppose that there is a homomorphism φ from Z 17 to some group, and that φ is not one-to-one. Determine φ. Ker(φ) = {e} if and only if φ is 1-1 Thus since φ is not 1-1, Ker(φ) {e}. Ker(φ) is a subgroup of Z17 which is cyclic of prime order. Since the only subgroups of Z 17 are the trivial subgroup or the whole group, Ker(φ) = Z 17. Thus every element in Z17 maps to the identity, and so φ : Z 17 G must be φ(n) = e for all n Z 17. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 4. If φ is a homomorphism from Z 30 onto a group of order 5, determine the kernel of φ. Let φ : Z 30 onto G, where G is a group of order 5. Then by the 1st isomorphism theorem, Z 30 /Ker(φ) φ(z 30 ) = G. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 4. If φ is a homomorphism from Z 30 onto a group of order 5, determine the kernel of φ. Let φ : Z 30 onto G, where G is a group of order 5. Then by the 1st isomorphism theorem, Z 30 /Ker(φ) φ(z 30 ) = G. Since Z 30 /Ker(φ) = Z 30 / Ker(φ) must equal G, we have that 30 = 5 = Ker(φ) = 6. Ker(φ) Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 4. If φ is a homomorphism from Z 30 onto a group of order 5, determine the kernel of φ. Let φ : Z 30 onto G, where G is a group of order 5. Then by the 1st isomorphism theorem, Z 30 /Ker(φ) φ(z 30 ) = G. Since Z 30 /Ker(φ) = Z 30 / Ker(φ) must equal G, we have that 30 = 5 = Ker(φ) = 6. Ker(φ) Because Z 30 is cyclic (of order 30), it has exactly one subgroup of order 6, so there s only one possibility for Ker(φ): Ker(φ) = 5 = {0, 5, 10, 15, 20, 25}. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Onto: Let φ : Z20 onto Z 10 be a homomorphism. Z 20 = 1 , so for all n Z 20, φ(n) = φ( mod 20) = φ(1)+... φ(1) mod 10 = nφ(1) mod 10. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Onto: Let φ : Z20 onto Z 10 be a homomorphism. Z 20 = 1 , so for all n Z 20, φ(n) = φ( mod 20) = φ(1)+... φ(1) mod 10 = nφ(1) mod 10. φ is onto k Z 10, n s.t. k = φ(n) = nφ(1) = φ(1) φ(1) Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Onto: Let φ : Z20 onto Z 10 be a homomorphism. Z 20 = 1 , so for all n Z 20, φ(n) = φ( mod 20) = φ(1)+... φ(1) mod 10 = nφ(1) mod 10. φ is onto k Z 10, n s.t. k = φ(n) = nφ(1) = φ(1) φ(1) Since every element of Z 10 can be generated by φ(1), φ(1) must be one of the generators of Z 10. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Onto: Let φ : Z20 onto Z 10 be a homomorphism. Z 20 = 1 , so for all n Z 20, φ(n) = φ( mod 20) = φ(1)+... φ(1) mod 10 = nφ(1) mod 10. φ is onto k Z 10, n s.t. k = φ(n) = nφ(1) = φ(1) φ(1) Since every element of Z 10 can be generated by φ(1), φ(1) must be one of the generators of Z 10. Thus we must have φ(1) U(10) = {1, 3, 7, 9}, so there are exactly 4 homomorphisms from Z 20 onto Z 10 : φ(n) = n φ(n) = 3n mod 10 φ(n) = 7n mod 10 φ(n) = 9n mod 10 Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Not necessarily onto: It is still true that φ(n) = nφ(1) for all n Z20, so it is still true that any homomorphism from Z 20 to Z 10 is completely defined by where φ sends 1. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Not necessarily onto: It is still true that φ(n) = nφ(1) for all n Z20, so it is still true that any homomorphism from Z 20 to Z 10 is completely defined by where φ sends 1. Because each homomorphism is completely defined by where φ sends 1, there s at most 1 homomorphism per destination in Z 10. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Not necessarily onto: It is still true that φ(n) = nφ(1) for all n Z20, so it is still true that any homomorphism from Z 20 to Z 10 is completely defined by where φ sends 1. Because each homomorphism is completely defined by where φ sends 1, there s at most 1 homomorphism per destination in Z 10. On the other hand, 1 can go anywhere in Z10 : we ll always end up with 0 mapping to 0, because 0 = 20 1 mod 20 in Z 20, so φ(0) = 20φ(1) mod 10 = 0. Thus there s at least 1 homomorphism per destination. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8 5. How many homomorphisms are there from Z 20 onto Z 10? How many are there to Z 10? (That is, how many are there that may or may not be onto?) Not necessarily onto: It is still true that φ(n) = nφ(1) for all n Z20, so it is still true that any homomorphism from Z 20 to Z 10 is completely defined by where φ sends 1. Because each homomorphism is completely defined by where φ sends 1, there s at most 1 homomorphism per destination in Z 10. On the other hand, 1 can go anywhere in Z10 : we ll always end up with 0 mapping to 0, because 0 = 20 1 mod 20 in Z 20, so φ(0) = 20φ(1) mod 10 = 0. Thus there s at least 1 homomorphism per destination. Thus there are 10 possible homomorphisms from Z20 into Z 10. Math 321-Abstract (Sklensky) In-Class Work November 29, / 8

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