Cosets of the subgroups

1. List the left cosets of the subgroups in each of the following. Here hai denotes the subgroup
   generated by the element a.
   (a) h8i in (Z24, +)
   (b) h3i in U(8)
   (c) 3Z in Z (where 3Z = {3k : k ∈ Z}.)
   (d) An in Sn (where An is the set of all even permutations on {1, . . . , n}.)
2. Find all the left cosets of H = {1, 19} in U(30).
3. Given a finite group G and H a subgroup. Using the same argument as in the Lemma proved
   in class, one can show all the statements of the lemma hold analogously for right cosets of H in G.
   In particular, Ha = H if and only if a ∈ H and the distinct right cosets of H form a partition of
   G. Now, suppose that |H| = |G|/2.
   (a) Show that for every a ∈ G, aH = Ha. (comment: in general aH and Ha are not necessarily
   equal. But with our condition |H| = |G|/2 here, this indeed holds.)
   (b) Suppose a, b ∈ G are two elements of G that are not in H. Prove that ab ∈ H.
4. Let G be a group of order 63. Prove that G must have an element of order 3.
5. Let G be a group of order 155. Suppose a, b are two nonidentity elements of G that have different
   orders. Prove that the only subgroup of G that contains both a and b must be G itself. (hint: By
   Larange’s theorem, the order of any nonidentity element must be one of 5, 31, 155. If one of a, b
   has order 155 then the statement is quite easy to prove. So one may assume |a| = 5 and |b| = 31.
   Consider how Theorem 7.2 might be relevant.)
6. Prove that every subgroup of Dn that has an odd order must be cyclic.

Sample Solution

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