- 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}.) - Find all the left cosets of H = {1, 19} in U(30).
- 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. - Let G be a group of order 63. Prove that G must have an element of order 3.
- 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.) - Prove that every subgroup of Dn that has an odd order must be cyclic.
Sample Solution
