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11, 5, and 13 have order 4, while 7, 9, and 15 have order 2, and 1 has order 1. Hence there is not element of order 8, so this group is not cyclic. (d) The powers of 2 mod 11 are 21 = 2, 22 = 4, 23 = 8, 24 = 5, 25 = 10, 26 = 9, 27 = 7, 28 = 3 29 = 6, 210 = 1. Thus, Z⁄ 11 is cyclic with generator 2. J 3. Find all of the generators of: (a) Z5 ...

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This is 99% correct, it's just that you've given subgroups of Z3 x Z9, not Z9 x Z3. Apr 9, 2006 #3 moo5003. 207 0. Wow... nice catch. Now I'm wondering if I can just write on top of the page to switch everything to the other side or if I should re-write the page :(.

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11 2, giving your answer as an element of Z 11. (iii) Calculate 3 × 13 12, ... Find two proper subgroups of G. Question 8 Find all the cyclic subgroups of Z 6. Analysis

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As a subgroup of the given group, <32> = <4> is cyclic of order 15, and a cyclic group of order 15 has precisely four subgroups. Last edited by Nehushtan (2016-03-16 07:32:57) 240 books currently added on Goodreads

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llenge. Previous studies have demonstrated the potential utility of epigenetic markers for identifying this group. Methods: A whole genome methylation interrogation using the Illumina HumanMethylation 450 array of patients with nondysplastic Barrett esophagus who either develop adenocarcinoma or remain static, with validation of findings by bisulfite pyrosequencing. Results: In all, 12 ...