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let g g be a cyclic group of order n 480 253151 h be a subgroup of g and m is the smallest natural number for which gm h f

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Let G = ⟨g⟩ be a cyclic group of order n = 480 = 253151, H be a subgroup of G, and m is the smallest natural number for which gm ∈ H. For which natural numbers m a) the index|G:H|ofGpoH sedelin2; b) the row|H|ofH sedelina2; c) the index|G:H|of G by H sedelina2 and sedelina3; d) the row|H|ofH sedline2 and isedline3.

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Answer to a math question Let G = ⟨g⟩ be a cyclic group of order n = 480 = 253151, H be a subgroup of G, and m is the smallest natural number for which gm ∈ H. For which natural numbers m a) the index|G:H|ofGpoH sedelin2; b) the row|H|ofH sedelina2; c) the index|G:H|of G by H sedelina2 and sedelina3; d) the row|H|ofH sedline2 and isedline3.

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98 Answers

a) To find the possible values of

m

for condition (a), we look for

m

such that

|H|

is a factor of 480 but not a power of 2. Given that

|H|

could be 15, 30, 60, 120, or 240, the values of

m

are 16, 32, or 64.

Answer:

m = 16, 32, 64

b) For condition (b), we need to find

m

such that

|H|

is a power of 2. Since

|H|

could be 1, 2, 4, 8, 16, 32, or 64, the possible values of

m

are 1, 2, 4, 8, 15, 30, or 60.

Answer:

m = 1, 2, 4, 8, 15, 30, 60

c) To satisfy condition (c), we look for

m

such that

|H|

is a divisor of 480 that leaves a quotient divisible by both 2 and 3. Given that

|H|

could be 80 or 240, the values of

m

satisfying this condition are 6 or 16.

Answer:

m = 6, 16

d) For condition (d), we need to find

m

such that

|H|

is divisible by both 2 and 3. Since

|H|

could be 3, 6, 12, 24, 48, 96, or 192, the possible values of

m

are 1, 2, 4, 5, 10, or 20.

Answer:

m = 1, 2, 4, 5, 10, 20

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