MAT-3110 Midterm: MATH 3110 App State Fall2011 Test3 answer key

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15 Feb 2019
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The order of an element of z9000 z3333333 is computed by nding the least common multiple of the orders of each of its components: |(a, b)| = lcm(|a|, |b|). If we have to have an element of order 3, then 3 = |(a, b)| = lcm(|a|, |b|). The only ways to get an lcm of 3 are lcm(3, 1) = lcm(1, 3) = lcm(3, 3) = 3. Now both z9000 and z3333333 are cyclic groups of orders divisible by 3. So they both have exactly 1 element of order. 1 (the identity) and 2 elements of order 3 (the two generators of the unique subgroup of order 3). 2 1 + 1 2 + 2 2 = 8 elements of order 3. Alternatively, taking all elements of order 1 or 3, we have 3 3 = 9 choices. Only the identity has order 1, so that leaves exactly 9 1 = 8 elements of order 3.

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