Math 546
Problem Set 17
1. Solve the puzzles on the peg-puzzle handout.
2. Recall that for any subgroup H of G, a ! b mod H " a #1b $H is an equivalence
relation on G. The equivalence class of an element a is the left coset aH .
(a). What is the equivalence class of the identity e?
Solution: H (i.e., [e] = eH = H )
(b). If a ! b mod H , how is aH related to bH?
Solution: aH = bH
(c). The notation a ! b mod H is meant to be suggestive of a ! b mod n . In fact if
we let H = nZ be the subgroup of Z consisting of all multiples of n, then
note that a ! b mod n is the same thing as a ! b mod H .
Solution: Just notice that.
(d). a ! b mod H and g is an element of G, do you think it must it be true
that ag ! bg mod H ?
Solution: You might think that, but it’s not necessarily so.
3. How many left cosets are there of K = {i,!(1,!2)} in S4 ? Solution: 12
(a). List the elements of the left coset (1,!2,!3)K .
Solution: (1,!2,!3)K = {(1,!2,!3)i,!(1,!2,!3)(1,!2)} = {(1,!2,!3),!(1,!3)} .
(b). List the elements of the right coset K(1,!2,!3) .
Solution: K(1,!2,!3) = {i(1,!2,!3),!(1,!2)(1,!2,!3)} = {(1,!2,!3),!(2,!3)} .
Are these two cosets the same?
Solution: Apparently not.
4. Let H =< (1,!2,!3,!4) > be the subgroup of S4 generated by the cycle (1,!2,!3,!4) .
(a). List the elements of H.
Solution: H = {(1,!2,!3,!4),!(1,!3)(2,!4),!(1,!4,!3,!2),!i}
(b). How many left cosets of H are there?
Solution: 6
(c). How many right cosets of H are there?
Solution: 6
(d). Determine the elements of the left and right cosets of H below.
(i). (1,!4)H
(ii). H (1,!4)
Solution:
(i). (1,!4)H = {(1,!2,!3),!!(1,!3,!4,!2),!!(2,!4,!3),!!1,!4)} .
(ii). H (1,!4) = {(2,!3,!4),!!(1,!2,!4,!3),!!(1,!3,!2),!!(1,!4)}
Are these two cosets the same? Solution: No they are not.
5. Recall that if A and B are subsets of a group G, then AB = {ab : a !A,!b !B} .
So, AB ! A B .
(a). Suppose that R = {e,!a},!!S = {e,!b} are subsets of the Klein 4-group.
What are the elements of RS?
Solution: RS = {ee,!eb,!ae,!ab} = {e,b, a, c} .
(b). Let A = {(1,!2),!(1,!3,!4)(2,!5)} and B = {(2,!3,!4),!!(1,!5,!2)} are subsets of S5 .
What are the elements of AB?
Solution: The elements of AB are:
(1,!2)(2,!3,!4) = (1,!2,!3,!4)
(1,!2)(1,!5,!2) = (1,!5)
(1,!3,!4)(2,!5)(2,!3,!4) = (1,!3)(2,!4,!5)
(1,!3,!4)(2,!5)(1,!5,!2) = (1,!2,!3,!4)
Hence, AB = {(1,!2,!3,!4),!!(1,!5),!!(1,!3)(2,!4,!5)} - just three elements.