MAT-3110 Midterm: MATH 3110 App State Fall2011 Test2 answer key

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15 Feb 2019
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Name: answer key: (20 points) cyclic (a) let g = hgi where g has order 20. Don"t forger g has order 20, so its exponents can be reduced modulo 20. hg8i = {g8, g16, g24, . = {g8, g16, g4, g12, e} = hg4i. Or g102 6 hg8i because gcd(8, 20) = 4 does not divide 102. (b) suppose g is a cyclic group with at least one element of order 6. Since g has an element of order 6, it"s order must be a multiple of 6. G could be a group of order 6, 12, 18, etc. If g g is an element of order 6, then hgi is a subgroup of order 6. Therefore, it has a unique subgroup of order 6. This subgroup in turn has unique subgroups of orders 1, 2, 3, and 6. Its subgroup of order 1 just contains the identity.

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