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Course 111: Algebra, 22nd November 2006 1. Consider the set of elements G = {0, 1, 2, 3, 4, 5} which forms a cyclic group under addition modulo 6. List the proper subgroups of G and the generators of the group. (Recall that {G} and {e}, where e = 0 here, are improper subgroups) Determine the left and right cosets of H = {0, 3} in G and construct the Cayley table of the quotient group G|H. Proper subgroups {0, 3} and {0, 2, 4}. Generators 1, 5 since all the group elements can be generated by repeated addition of 1 to itself (mod 6) or 5 to itself (mod 6). Left & right cosets of H = {0, 3} in G {0, 3}, {1, 4}, {2, 5} and the left and right cosets are the same. Cayley table The group G|H is the set of left (or right) cosets = {{0, 3}, {1, 4}, {2, 5}} and writing {0, 3} = H, {1, 4} = A, {2, 5} = B the Cayley table is H A B H H A B A B A B B H H A 2. Recall the properties of cosets listed in your notes. From this list prove the following: For a subgroup H of a group G, • x ∈ xH for all x ∈ G. Proof: Let x ∈ G. Then x = xe for e the identity element of G. But also, e ∈ H. Then it follows that x ∈ xH. • if x and y ∈ G and if y = xa for some a ∈ H then xH = yH. Proof: Let x, y ∈ G with y = xa for some element a ∈ H. Then yh = x(ah) and xh = y(a−1 h) ∀h ∈ H. Also, ah ∈ H and a−1 h ∈ H ∀h ∈ H because H a subgroup of G. Therefore, yH ⊂ xH and xH ⊂ yH and so yH = xH. • each left coset of H in G has the same number of elements as H. Proof: Consider xH ∩ yH to be nonempty for some x, y ∈ G. And z ∈ (xH ∩ yH). Then z = xa for some a ∈ H and z = yb for some b ∈ H. Then from the second proof here, zH = xH and zH = yH so xH = yH.