2. Consider the group U(13) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} with multiplication modulo 13. Let H = {1, 3, 9}. This is a normal subgroup of U(13) (you don't need to write up a proof of that, but make sure you know how to prove it!). (a) How many distinct cosets of H in U(13) are there? (b) List the distinct cosets of H in U(13). These are the elements of the quotient group U(13)/H. (c) Create a Cayley table for the quotient group U(13)/H. (d) What is the inverse of each element in U(13)/H? (e) What is the order of each element in U(13)/H? (f) Is the group U(13)/H abelian? Briefly explain your answer. (g) Is the group U(13)/H cyclic? Briefly explain your answer. (h) Depending on your answer to the previous question, prove that U(13)/H is isomorphic to Z4 or Z2 x Z2.
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In this case, U(13) has a multiplication modulus of 13. So, to generate the cosets of U(13), we need to multiply each element in U(13) by 13. Show more…
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