S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
Some problems are really easy, some are harder, some are repetitions.
1. Let Sn be the group of permutations on n elements {1, 2, 3, . . . , n}. Let An be the subgroup of even permutations. Prove that An is normal subgroup of Sn. Answer:
• One way:
◦ |Sn| = n! and |An| = n!/2
◦ # (left cosets of An in Sn)= [Sn : An] = |Sn|/|An| = n!/(n!/2) = 2
◦ # (right cosets of An in Sn)=# (left cosets of An in Sn)=2
◦ Want to show: aAn = Ana for all a ∈ Sn - If a ∈ An then aAn = An = Ana. - If a∈ / An then aAn 6= An 6= Ana Sn = An ∪ aAn disjoint union of 2 distinct left cosets. Sn = An ∪ Ana disjoint union of 2 distinct right cosets. An ∪ aAn = An ∪ Ana disjoint unions implies aAn = Ana for all a∈ / An
◦ An is normal subgroup in Sn since aAn = Ana for all a ∈ Sn • Another way:
◦ An = {even permutations on n elements} ◦ A subgroup H in a group G is normal subgroup, if and only if gHg−1 ⊂ H. Propo- sition proved in class. −1 ◦ WTS σAnσ ⊂ An Let α ∈ An. Then α is an even permutation (by definition of An). −1 Then σασ is even permutation for all σ ∈ Sn (by a problem from a previous quiz). −1 Therefore σασ ∈ An −1 Therefore σAnσ ⊂ An Therefore An is normal subgroup in Sn
2. Let S15 be the group of permutations on 15 elements {1, 2, 3,..., 15}. Prove that S15 has elements of order 56 but does not have any elements of orders 49 or 50. Answer: Use the following facts:
(a) Order of a k-cycle is k, i.e. |α| = k for α a k-cycle. (b) |αβ| = lcm(|α|, |β|) if α and β are disjoint permutations.
• S15 has elements of order 56: Let σ = αβ, where α is an 8-cycle and β is a 7-cycle. Then |σ| = |αβ| = lcm(|α|, |β|) = lcm(8, 7) = 56.
• S15 does not have any elements of orders 49: In order for lcm(m, n) = 49, either m = 49 or n = 49. Therefore we would need a cycle of order 49, hence need a 49-cycle in S15, which is impossible.
1 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
• S15 does not have any elements of orders 50: In order for lcm(m, n) = 50, either m or n 2 must be a multiple of 5 = 25. Therefore we would need a 25-cycle or 50-cycle in S15, which is impossible. ( Factor Groups - Quotient Groups - G/H )
3. Prove that a factor group of a cyclic group is cyclic. Answer: Recall: A group G is cyclic if it can be generated by one element, i.e. if there exists an element a ∈ G such that G =< a > (this means that all elements of G are of the form ai for some integer i.) Recall: Elements of a factor group G/H are left cosets {gH | g ∈ G. Proof: Suppose G =< a >. Let G/H be any factor group of G. WTS: G/H is cyclic. Any element of G/H is of the form gH for some g ∈ G. Since G is cyclic, there is an integer i such that g = ai. So gH = aiH. aiH = (aH)i by definition of multiplication in factor groups. Therefore gH = (aH)i for any coset gH. So G/H is generated by aH, hence G/H is cyclic, by the definition of cyclic groups.
4. Prove that a factor group of an Abelian group is Abelian. Answer: Recall: A group G is Abelian if ab = ba for all a, b ∈ G. Proof: Suppose G is Abelian. Let G/H be any factor group of G. WTS: G/H is Abelian. Let aH, bH be any elements of G/H. Then (aH)(bH) = (ab)H by definition of multiplication in factor groups. Then (ab)H = (ba)H since G is abelian. (ba)H = (bH)(aH) again definition of multiplication in factor group. Therefore (aH)(bH) = (bH)(aH). Therefore G/H is Abelian.
5. Prove that any subgroup of an Abelian group is normal subgroup. Answer: Recall: A subgroup H of a group G is called normal if gH = Hg for every g ∈ G. Proof: Suppose G is Abelian. Let H be a subgroup of G. WTS: gH = Hg for all g ∈ G. Let g ∈ G. Then gH = {gh | h ∈ H} by definition of left coset. gh = hg for all h since G is Abelian. Therefore {gh | h ∈ H} = {hg | h ∈ H} = Hg by definition of right coset Hg. Therefore gH = Hg for all g ∈ G. Therefore H is normal subgroup of G.
2 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
6. Let H =< 5 > be the subgroup of G = Z generated by 5. (a) List the elements of H =< 5 >. Answer: Notice: Operation in G = Z is addition. H = h5i = {0, ±5, ±10,... } = {5k | k ∈ Z} (b) Prove that H is normal subgroup of G. Answer: Addition in Z is commutative. So G = (Z, +) is Abelian group and by previous problem every subgroup of an Abelian group is normal. (c) List the elements of the factor group G/H = Z/ < 5 >. Answer: Elements of G/H are left cosets of H in G: 0 + H = 0 + h5i = {0 + 0, 0 ± 5, 0 ± 10,... } = {0, ±5, ±10,... } = H 1 + H = 1 + h5i = {1 + 0, 1 ± 5, 1 ± 10,... } = {..., −9, −4, 1, 6, 11,... } 2 + H = 2 + h5i = {2 + 0, 2 ± 5, 2 ± 10,... } = {..., −8, −3, 2, 7, 12,... } 3 + H = 3 + h5i = {3 + 0, 3 ± 5, 3 ± 10,... } = {..., −7, −2, 3, 8, 13,... } 4 + H = 4 + h5i = {4 + 0, 4 ± 5, 4 ± 10,... } = {..., −6, −1, 4, 9, 14,... } (d) Write the Cayley table of G/H = Z/ h5i.
G/H = Z/ h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 1 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 2 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 1 + h5i 3 + h5i 3 + h5i 4 + h5i 0 + h5i 1 + h5i 2 + h5i 4 + h5i 4 + h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i
(e) What is the order of 2 + h5i in Z/ h5i? Answer: |2 + h5i | = 5 as 2 + h5i ∈ Z/ h5i.
3 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
7. Let K = h15i be the subgroup of G = Z generated by 15. (a) List the elements of K = h15i. Answer: K = h15i = {15k | k ∈ Z} (b) Prove that K is normal subgroup of G. Proof: (Z +) is Abelian group and any subgroup of an Abelian group is normal (from 5). (c) List the elements of the factor group G/K = Z/ h15i. Answer: G/K = Z/ h15i = {i + h15i | 0 ≤ i ≤ 14}. (There are 15 elements.) !!! This is just one way of expressing these cosets. Notice that there are many ways of expressing the same coset, eg. (2+h15i) = (17+h15i) = (32+h15i) = (−13+h15i) = ... . (d) Write the Cayley table of G/K = Z/ h15i. Answer: Don’t have to actually write it, but make sure that you know what it would look like. (e) What is the order of 3 + K = 3 + h15i in Z/ h15i? Answer: |3 + h15i | = 5 in Z/ h15i, since (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) = (15 + h15i) = (0 + h15i). (f) What is the order of 4 + K = 4 + h15i in Z/ h15i? Answer: |4 + h15i | = 15 in Z/ h15i. (g) What is the order of 5 + K = 5 + h15i in Z/ h15i? Answer: |5 + h15i | = 3 in Z/ h15i. (h) What is the order of 6 + K = 6 + h15i in Z/ h15i? Answer: |6 + h15i | = 5 in Z/ h15i, (i) Prove that G/K is cyclic. Answer: Z/ h15i is generated by 1 + h15i, hence it is cyclic. underlineAnswer 2: Z/ h15i is cyclic since it is factor of the cyclic group (Z, +) (this group is generated by 1).
(j) Prove that G/K = Z/ h15i is isomorphic to Z15. Answer: • One way - using the First Isomorphism Theorem:
– Define a group homomorphism: f : Z → Z15: ∗ Since Z is cyclic it is enough to define homomorphism on a generator, and extend to all other elements. ∗ Define f(1) := 1(mod15) and f(n1) := n1 (mod15) (i.e. f(n) := n (mod15) ∗ Since |1| = ∞ for 1 ∈ Z and |f(1)| = 15 for f(1) = 1 ∈ Z/ h15i we have |f(1)| | |1|. – Claim 1: f is onto (this is quite clear, but here is a detailed proof).
4 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
∗ Elements of Z15 are integers {0, 1, 2,..., 14} ∗ For each n ∈ {0, 1, 2,..., 14} = Z15 consider n ∈ Z. ∗ Then f(n) = n ∈ Z15. Hence f is onto. – Im(f) = Z15 (As stated in class this is equivalent to f being onto.) – Claim 2: Ker(f) = h15i < Z (this is quite clear, but here is a detailed proof). ∗ Ker(f) = {n ∈ Z | f(n) = 0 ∈ Z15} ∗ n (mod15) = r ∈ {0, 1,..., 14}, where r is the remainder after dividing n by 15, i.e. n = 15k + r, for some integer k. ∗ If n ∈ Ker(f) then f(n) = 0 and therefore n = 15k ∈ h15i < Z. ∗ Therefore Ker(f) ⊂ h15i. ∗ h15i ⊂ Ker(f) is clear since x ∈ h15i implies x = 15k for some integer k and therefore f(x) = f(15k) = 15k(mod15) = 0inZ15. – First Isomorphism Theorem states: If f : G → G0 is a group homomorphism then G/Ker(f) ∼= Im(f). ∼ – Z/ h15i = Z15 • Another way, by defining isomorphism and checking all details of being isomorphism: – Elements of Z/ h15i are left cosets, and operation is addition of cosets (as defined for factor groups!)
– Elements of Z15: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} and addition is modulo 15.
– Define the map: f : Z/ h15i → Z15 by f(i + h15i) = ri, where ri is the remainder after dividing i by 15.
∗ Notice that f, as defined, is a mapping from Z/ h15i to Z15 since the values of f are the remainders after dividing by 15, hence they are non-negative integers between 0 and 14. ∗ Prove that f is well-defined: Suppose the same coset (i + h15i) is given by another integer j, i.e. (i + h15i) = (j + h15i). Then f(i + h15i) = ri ∈ Z15 and f(j + h15i) = rj ∈ Z15. To show that f is well defined we must show that f(i + h15i) = f(j + h15i), i.e. WTS ri = rj ∈ Z15. From (i + h15i) = (j + h15i) it follows that j ∈ (i + h15i) and therefore j = i + k15 for some integer k ∈ Z. By definition of remainders: i = n · 15 + ri and j = m · 15 + rj, with 0 ≤ ri, rj < 15. Therefore from j = i + k15, we have m · 15 + rj = n · 15 + ri + k15. Hence rj −ri = (k+n−m)·15 which is an integer multiple of 15. Since the remainders are 0 ≤ ri, rj < 15, it follows that rj = ri. Therefore f is a well defined function.
5 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
∗ Let f : Z/ h15i → Z15 defined by f(i + h15i) = ri, where ri is the remainder after dividing i by 15. Then f is ”one-to-one” function. Proof: Suppose there are a, b ∈ Z/ h15i such that f(a) = f(b). WTS a = b. From a, b ∈ Z/ h15i, it follows that a = i + h15i and b = j + h15i. From f(a) = f(b) it follows f(i + h15i) = f(j + h15i), and by definition of f it follows that ri = rj. So i = n · 15 + ri and j = m · 15 + rj. Therefore i − j is a multiple of 15, hence and element of the subgroup h15i. Therefore a = i + h15i = j + h15i = b. Therefore f is a one-to-one function.
∗ f : Z/ h15i → Z15 is onto. Proof: Let y ∈ Z15. WTS: there is an x ∈ Z/ h15i such that f(x) = y. Since y ∈ Z15, y is an integer 0 ≤ y < 15. Let x = y + h15i ∈ Z/ h15i. Then f(x) = f(y + h15i) = ry. Since 0 ≤ y < 15, it follows that ry = y. Therefore f(x) = y. Therefore f is onto. ∗ f(a + b) = f(a) + f(b) Proof: Let a, b ∈ Z/ h15i. From a, b ∈ Z/ h15i, it follows that a = i + h15i and b = j + h15i. Then it follows that a + b = (i + h15i) + (j + h15i) = (i + j) + h15i f(a + b) = (i + j)(mod15) f(a) + f(b) = i (mod15) + j (mod15) = (i + j)(mod15) Therefore f(a + b) = f(a) + f(b)
– Therefore f : Z/ h15i → Z15 is an isomorphism.
6 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
8. Let G = h6i and H = h24i be subgroups of Z. Show that H is a normal subgroup of G. Write the cosets of H in G. Write the Cayley table for G/H.
Answer: G and H are subgroups of Z, hence operation is addition. G = h6i = {0, ±6, ±12, ±18, ±24, ±30, ±36, ±42, ±48,... } = {6j | j ∈ Z}, multiples of 6. H = h24i = {0, ±24, ±48,... } = {24j | j ∈ Z}, i.e. all integer multiples of 24. • H is a normal subgroup of G. – H is a nonempty subset of G, since elements of H are elements of G (integer multiples of 24 are multiples of 6). – H is closed under operation: If a, b ∈ H, then a = 24m, and b = 24n. Therefore a + b = 24m + 24n = 24(m + n) ∈ H. – H is closed under inverses: If a ∈ H, then a = 24m. Therefore −a = 24 · (−m), hence −a ∈ H. – Since Z is abelian, G is also abelian and therefore any subgroup of G is normal. Hence H is normal in G. • Cosets of H = h24i in G = h6i are: H = 0 + H = 0 + h24i = {..., −48, −24, 0, 24, 48,... }, 6 + H = 6 + h24i = {..., −42, −18, 6, 30, 54,... }, 12 + H = 12 + h24i = {..., −36, −12, 12, 36, 60,... }, 18 + H = 18 + h24i = {..., −30, −6, 18, 42, 66,... }. • Elements of G/H are the four cosets written above and the Cayley table is:
G/H = h6i / h24i 0 + h24i 6 + h24i 12 + h24i 18 + h24i 0 + h24i 0 + h24i 6 + h24i 12 + h24i 18 + h24i 6 + h24i 6 + h24i 12 + h24i 18 + h24i 0 + h24i 12 + h24i 12 + h24i 18 + h24i 0 + h24i 6 + h24i 18 + h24i 18 + h24i 0 + h24i 6 + h24i 12 + h24i
9. Viewing h6i and h24i as subgroups of Z, prove that h6i/h24i is isomorphic to Z4. Proof: Elements of h6i/h24i are the 4 cosets: {(0 + h24i), (6 + h24i), (12 + h24i), (18 + h24i)} with the above multiplication table. |6 + h24i | = 4 since (6 + h24i) + (6 + h24i) + (6 + h24i) + (6 + h24i) = (0 + h24i). So h6i/h24i can be generated by one element of order 4, therefore it is cyclic of order 4.
The group Z4 is also cyclic of order 4. By theorem: ”Any two cyclic groups of the same order are isomorphic.”, it follows that h6i/h24i is isomorphic to Z4.
7 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
10. Let h8i be the subgroup of Z48.
(a) What is the order of the factor group Z48/h8i? Answer: Elements of Z48 are {0, 1, 2, 3,... 46, 47} and |Z48| = 48. Elements of h8i ⊂ Z48 are {0, 8, 16, 24, 32, 40} and | h8i | = 6. Therefore the order of the factor group is: |Z48/h8i | = |Z48|/| h8i | = 48/6 = 8.
(b) What is the order of 2 + h8i in the factor group Z48/h8i? Answer: Elements of Z48/ h8i are the following cosets: {(0 + h8i), (1 + h8i), (2 + h8i), (3 + h8i), (4 + h8i), (5 + h8i), (6 + h8i), (7 + h8i)}. The order of (2 + h8i) is |(2 + h8i | = 4 since 4 is the smallest number of times (2 + h8i) must be added to itself in order to get the identity, i.e. such that (2 + h8i) + (2 + h8i) + (2 + h8i) + (2 + h8i) = (0 + h8i). 11. Let G = U(16) be the group of units modulo 16. (a) What is the order of G? Answer: G = {1, 3, 5, 7, 9, 11, 13, 15}. So |G| = 8. We also know that |U(16)| = φ(16) = φ(24) = 23(2 − 1) = 8. (b) What is the order of 15 ∈ U(16)? Answer: 15 · 15 = 225 ≡ 1(mod16). Therefore |15| = 2 in U(16). (Another way: 15 · 15 = (−1)(−1) = 1 ≡ 1(mod16)) (c) Let H =< 15 > be the subgroup of U(16) generated by 15. What is the order of the factor group U(16)/H? Answer: |G| = 8, |H| = 2. Therefore |G/H| = |G|/|H| = 8/2 = 4 Also, in more details: H =< 15 >= {1, 15}. Elements of G/H are the left cosets of H in G: 1H = {1, 15} = 15H, 3H = {3, 13} = 13H 5H = {5, 11} = 11H 7H = {7, 9} = 9H (d) Make the Cayley table of the factor group U(16)/H. Answer:
G/H 1H 3H 5H 7H 1H 1H 3H 5H 7H 3H 3H 7H 1H 5H 5H 5H 1H 7H 3H 7H 7H 5H 3H 1H
8 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
( Group Homomorphisms )
12. Let F : Z → Z be a function defined as F (x) = 10x. (a) Prove that F is a group homomorphism. Proof: • F is a function F : Z → Z (already stated in the problem, so no need to prove it). • F (a + b) = 10(a + b) by definition of F . 10(a + b) = 10a + 10b distributive property of integers. F (a) + F (b) = 10a + 10b by definition of F . Therefore F (a + b) = F (a) + F (b). • Therefore F is a group homomorphism. (b) Find Ker(F ). Answer: • Ker(F ) = {x ∈ Z | F (x) = 0 ∈ Z} = {x ∈ Z | 10x = 0 ∈ Z} = {0}. • Ker(F ) = {0} (c) Find Im(F ) Answer: • Im(F ) = {y ∈Z | there is an x∈Z such that F (x)=y} = {10x∈Z | x ∈ Z} = 10Z. • Im(F ) = h10i < Z
13. Let f : Z6 → D3 be the map defined as f(1) = ρ. (a) Find f(2) f(2) = f(1 + 1) = f(1)f(1) = ρρ = ρ2 (b) Find f(3) 3 f(3) = f(1 + 1 + 1) = f(1)f(1)f(1) = ρρρ = ρ = e ∈ D3 (c) Find f(4) f(4) = f(4 · 1) = f(1)4 = ρ4 = ρ (d) Find f(5) f(5) = f(1)5 = ρ5 = ρ2 (e) Find f(0) f(0) = e (f) Find Ker(f) Ker(f) = {0, 3} = h3i < Z6 (g) Find Im(f) 2 Im(f) = {e, ρ, ρ } = hρi < D3
(h) Find Z6/Ker(f) Z6/Ker(f) = {Ker(f), 1+Ker(f), 2+Ker(f)} = {h3i , 1+h3i , 2+h3i}
9 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
(i) Describe an isomorphism Z6/Ker(f) → Im(f) Define ϕ : Z6/Ker(f) → Im(f) Define ϕ : Z6/ h3i → hρi < D3 by ϕ(1+h3i) = ρ and therefore ϕ(2+h3i) = ρ2 and ϕ(h3i)) = ϕ(0+h3i) = ϕ(3+h3i) = ρ3 = e.
14. Let f : G → G0 be a group homomorphism. Prove: If Ker(f) = e then f is a one-to-one map.
15. Let f : G → G0 be a group homomorphism. Prove: Ker(f) is a normal subgroup of G.
16. If |G| = 18 and a ∈ G, what are all possible orders of a? Answer:
• Order of each element of a group divides order of the group, i.e. |a| | |G| • |a| = 1, 2, 3, 6, 9, 18
17. If |G| = 17 and a ∈ G, what are all possible orders of a? Answer:
• Order of each element of a group divides order of the group, i.e. |a| | |G| • |a| = 1, 17
18. If |G| = 23 and a ∈ G, a 6= e, what are all possible orders of a? Answer:
• Order of each element of a group divides order of the group, i.e. |a| | |G| • |a| = 1, 23 • |a| =6= 1 since a 6= e and e is the only element in G od order 1. • |a| = 23
19. Let f : Z15 → Z12 be a group homomorphism defined by f(x) = 4x(mod 12). • Find f(5). f(5) = 4 · 5 (mod12) = 20 (mod12) = 8 ∈ Z12 • Find f(6). f(6) = 4 · 6 (mod12) = 24 (mod12) = 0 ∈ Z12 • Find Im(f). Im(f) = {4, 8, 0} = h4i < Z12 • Find Ker(f). Ker(f) = {3, 6, 9, 12, 0} = h3i < Z15 • Is f onto? No, since Im(f) 6= Z12
10 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
• Is f one-to-one? No, since Ker(f) 6= {0}. Also, easy to see: there are 5 different elements which are all mapped to the same element. • Is f an isomorphism? No, f is neither one-to-one, not onto.
20. Let f : Z15 → Z25 be a group homomorphism. What are all possible orders of f(1)?
• |1| = 15 in Z15 • |f(a)| | |a| implies that |f(1)| = 1, 3, 5, 15. • |g| | |G| implies that f(1)|25 and therefore f(1) = 1, 5, 25 • Therefore: |f(1)| = 1, 5.
21. Describe all possible group homomorphisms Z15 → Z25.
• Since Z15 is cyclic, it is enough to define homomorphism on 1 ∈ Z15, and extend to other elements.
• The only possibilities for the image of 1 ∈ Z15 are elements of order 1 or 5. • Case 1: |f(1)| = 1. This implies: f(1) = 0 and therefore for all x ∈ Z15, f(x) = 0 ∈ Z25. • Case 2: |f(1)| = 5. This implies: f(1) = 5, 10, 15, 20, i.e. all integers 1 ≤ i ≤ 25 such that gcd(i, 25) = gcd(5, 25) = 5. So, there are 4 such homomorphisms: f1(1) = 5 and therefore for all x ∈ Z15, f1(x) = 5x ∈ Z25, f2(1) = 10 and therefore for all x ∈ Z15, f2(x) = 10x ∈ Z25, f3(1) = 15 and therefore for all x ∈ Z15, f3(x) = 15x ∈ Z25, f4(1) = 20 and therefore for all x ∈ Z15, f4(x) = 20x ∈ Z25.
22. Describe all possible group isomorphisms Z15 → Z25. • There are no isomorphisms, since the groups have different orders: |Z15| = 15 and |Z25| = 25 and therefore cannot be isomorphic.
23. Describe all possible group isomorphisms Z25 → Z25.
• Since Z25 is cyclic, it is enough to define homomorphism on 1 ∈ Z25, and extend to other elements. • In order for the homomorphism to be isomorphism the order of f(1) must be 25.
• The elements of order 25 in Z25 are 1 ≤ i ≤ 25 such that gcd(i, 25) = gcd(1, 25) = 1. • The elements of order 25 in Z25 are {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24} • Also the number of elements in Z25 relatively prime to 25 is equal to Euler function ϕ(25).
11 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
2 • ϕ(25) = ϕ(5 ) = 5(5 − 1) = 20. Therefore, there are 20 isomorphisms Z25 → Z25: f1(x)=1x, f2(x)=2x, f3(x)=3x, f4(x)=4x, f5(x)=6x, f6(x)=7x, f7(x)=8x, f8(x)=9x, f9(x)=11x, f10(x)=12x, f11(x)=13x, f12(x)=14x, f13(x)=16x, f14(x)=17x, f15(x)=18x, f16(x)=19x, f17(x)=21x, f18(x)=22x, f19(x)=23x, f20(x)=24x
24. How many group isomorphisms Z75 → Z75.
• (# group isomorphism Z75 → Z75)= (# generators of Z75)= (# of integers i, such that 1 ≤ i ≤ 75 and gcd(i, 75) = 1) = ϕ(75) = ϕ(3 · 52) = ϕ(3)ϕ(52) = (3 − 1)(5)(5 − 1) = (2)(5)(4) = 40
• There are 40 group isomorphisms Z75 → Z75.
25. Prove that there is no group homomorphism f : Z10 → Z25 such that f(1) = 3.
• |1| = 10 in Z10. Therefore |f(1)| = 1, 2, 5, 10
• gcd(3, 25) = 1 implies that |3| = 25 in Z25. • Therefore f(1) 6= 3.
26. Prove that there is exactly one group homomorphism f : Z10 → Z49.
• |1| = 10 in Z10. Therefore |f(1)| = 1, 2, 5, 10
• f(1) ∈ Z49 implies |f(1)| = 1, 7, 49 • |f(1)| = 1 form the above two statements.
• f(x) = 0 ∈ Z49 for all x ∈ Z10.
27. Prove that there is exactly one group homomorphism f : Z11 → S6.
• |1| = 11 in Z11. Therefore |f(1)| = 1, 11
• |S6| = 6! implies that order of elements of S6 must divide 6!. • Since 11 does not divide 6!, the order of f(1) must be 1.
• f(x) = e ∈ S6 for all x ∈ Z11
28. Describe all group homomorphisms Z9 → S3. In each case describe Image and Kernel of the homomorphism.
• |1| = 9 in Z9. Therefore |f(1)| = 1, 3, 9
• S3 has elements of order 1,2,3. • The possible orders for f(1) are 1,3, and the possible functions:
– f1 : Z9 → S3 is defined as f1(i) = (1)
12 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
∗ Im(f1) = {e} = {(1)} = h(1)i < S3 (These are just different forms of writing the same subgroup.)
∗ Ker(f1) = Z9 – f2(1)=(123), and extend to all i ∈ Z9: i ∗ f2 : Z9 → S3 is defined as f2(i) = (123) 2 4 ∗ Computtaions of f2(i): f2(2)=(123) =(132), f2(3)=(1), f2(4)=(123) =(123), 5 6 7 f2(5)=(123) =(132), f2(6)=(123) =(1), f2(7)=(123) =(123), 8 9 f2(8)=(123) =(132), f2(0)=f2(9)=(123) =(1),
∗ Im(f2) = {(123), (132), (1)} = h(123)i < S3 ∗ Ker(f2) = {3, 6, 0} = h3i < Z9 – f3(1)=(132), and extend to all i ∈ Z9: i ∗ f3 : Z9 → S3 is defined as f3(i) = (132)
∗ Im(f3) = {(132), (123), (1)} = h(132)i < S3 ∗ Ker(f3) = {3, 6, 0} = h3i < Z9
29. Describe all group homomorphisms Z9 → S4. In each case describe Image and Kernel of the homomorphism.
• Possible orders for f(1) are 1 or 3.
• f1(x) = e, for all x ∈ Z9 Ker(f1) = Z9 Im(f1) = h(123)i x • f2(x) = (123) , for all x ∈ Z9 Ker(f2) = h3i < Z9 Im(f2) = h(123)i x • f3(x) = (132) , for all x ∈ Z9 Ker(f3) = h3i < Z9 Im(f3) = h(132)i x • f4(x) = (124) , for all x ∈ Z9 Ker(f4) = h3i < Z9 Im(f4) = h(124)i x • f5(x) = (142) , for all x ∈ Z9 Ker(f5) = h3i < Z9 Im(f5) = h(142)i x • f6(x) = (134) , for all x ∈ Z9 Ker(f6) = h3i < Z9 Im(f6) = h(134)i
13 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
x • f7(x) = (143) , for all x ∈ Z9 Ker(f7) = h3i < Z9 Im(f7) = h(143)i x • f8(x) = (234) , for all x ∈ Z9 Ker(f8) = h3i < Z9 Im(f8) = h(234)i x • f9(x) = (243) , for all x ∈ Z9 Ker(f9) = h3i < Z9 Im(f9) = h(243)i
30. Describe all group homomorphisms Z9 → U(10). In each case describe Image and Kernel of the homomorphism.
• Order of 1 ∈ Z9 is 9. Therefore |f(1)| = 1, 3, 9. • U(10) = {1, 3, 7, 9}, therefore |U(10)| = 4. So |f(1)| = 1, 2, 4
• Therefore |f(1)| = 1. Hence f(1) = 1 ∈ U(10). Extend to all i ∈ Z9. i • f(i · 1) = 1 = 1 ∈ U(10) for each i ∈ Z9. • Im(f) = {1} = h1i < U(10).
• Ker(f) = Z9.
31. Describe all group homomorphisms Z9 → U(9). In each case describe Image and Kernel of the homomorphism.
• Order of 1 ∈ Z9 is 9. Therefore |f(1)| = 1, 3, 9. • U(9) = {1, 2, 4, 5, 7, 8}, therefore |U(9)| = 6. So |f(1)| = 1, 2, 3, 6 • Therefore |f(1)| = 1, 3. • Case 1. |f(1)| = 1.
– Then f(1) = 1 ∈ U(9). Extend to all i ∈ Z9 in order to get: f : Z9 → U(9) i – f(i · 1) = 1 = 1 ∈ U(9) for each i ∈ Z9. – Im(f) = {1} = h1i < U(9). – Ker(f) = Z9. • Case 2. |f(1)| = 3. – First find orders of each of the elements in U(9): |1| = 1 Since order of the identity in any group is always 1. |2| = 6 Since 21 = 2, 22 = 4, 23 = 8, 24 = 7, 25 = 5, 26 = 1 ∈ U(9) |4| = 3 Since 41 = 4, 42 = 7, 43 = 1 ∈ U(9) |5| = 6 Since 51 = 5, 52 = 7, 53 = 8, 54 = 4, 55 = 2, 56 = 1 ∈ U(9) |7| = 3 Since 71 = 7, 72 = 4, 73 = 1 ∈ U(9) |8| = 2 Since 81 = 8, 82 = 1 ∈ U(9)
14 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name:
– So, there are two more homomorphisms: f2(1) = 4 and f3(1) = 7
– f2(1) = 4 ∈ U(9).
∗ Extend to all i ∈ Z9 in order to get: f2 : Z9 → U(9) i ∗ f2(i · 1) = 4 ∈ U(9) for each i ∈ Z9. 2 3 4 5 ∗ f2(1)=4, f2(2)=4 =7, f2(3)=4 =1, f2(4)=4 =4, f2(5)=4 =7, 6 7 8 9 f2(6)=4 =1, f2(7)=4 =4, f2(8)=4 =7, f2(0)== f2(9)=4 =1
∗ Im(f2) = {4, 7, 1} = h4i < U(9). ∗ Ker(f2) = {3, 6, 0} = h3i < Z9. – f3(1) = 7 ∈ U(9).
∗ Extend to all i ∈ Z9 in order to get: f3 : Z9 → U(9) i ∗ f3(i · 1) = 7 ∈ U(9) for each i ∈ Z9. 2 3 4 5 ∗ f3(1)=7, f3(2)=7 =4, f3(3)=7 =1, f3(4)=7 =7, f3(5)=7 =4, 6 7 8 9 f3(6)=7 =1, f3(7)=7 =7, f3(8)=7 =4, f3(0)== f3(9)=7 =1
∗ Im(f3) = {4, 7, 1} = h4i < U(9). ∗ Ker(f3) = {3, 6, 0} = h3i < Z9.
15