(1) (5 points) (a)What is the largest order of an element in S5? The largest order
is 6 ... Now, note that in general (bab−1)n = (bab−1)(bab−1) ···(bab−1) = banb−1.
Homework-Assignment 5 Name: Selected solutions. (1) (5 points) (a)What is the largest order of an element in S5 ? The largest order is 6 and a permutation of order is 6 is (12)(345). (b)Let G = Z2 × Z4 . Find an element of order 2 in G. The element (1, ˆ 0) has the property (1, ˆ 0) + (1, ˆ 0) = (0, ˆ 0) so it has order 2. (c)Show that if H, K are subgroups of a group G, then H ∩ K is a subgroup of G. We need to check that ∀x, y ∈ H ∩ K, xy −1 ∈ H ∩ K. But we know that H and K are subgroup so xy −1 ∈ H and xy −1 ∈ K. (d)What is is the order of 1 1 A=( ) 0 1 in GL2 (R)? Check that 1 n ) 0 1 so this never equals the identity matrix. Hence the order is infinite. (e)Let G be a group and a ∈ G. Prove that ord(a) = ord(a−1 ). Note that ord(a) equals the cardinality of (a). Similarly, ord(a−1) equals the cardinality of −1 (a ). But (a) = {an : n ∈ Z} = (a−1 ) and this ends the proof. (2) (5 points) Let a, b two elements in a group G. Prove that ord(a) = ord(bab−1). An = (
Proof. We need to show that min{n > 0 : an = 1} = min{n > 0 : (bab−1)n = 1}. Therefore it is enough to show that {n > 0 : an = 1} = {n > 0 : (bab−1)n = 1}. Now, note that in general (bab−1)n = (bab−1)(bab−1) · · · (bab−1) = ban b−1. Let us show that an = 1 if and only if (bab−1)n = 1. Assume that an = 1. Then (bab−1)n = ban b−1 = b · 1 · b−1 = 1. Conversely, assume that (bab−1)n = 1. Then ban b−1 = 1 so, by multiplying on the left with b−1 we get an b−1 = b−1 . Multiplying with b on the right we get an = 1. This establishes what we wanted to show. (3) (5 points) Find all the permutations σ in S6 with the property σ 4 = e. (4) (5 points) Find the center of D4 . (5) (5 points)(for graduate students only) Let H be a subgroup of G. Prove that there is a one-toone correspondence between the set of left cosets of H in G and the set of right cosets of H in G. Proof. Let L = set of left cosets, R=set of right cosets. Let F : L → R defined by F (aH) = Ha−1 . Remember that Hx = Hy if and only if xy −1 ∈ H. Also, xH = yH if and only if x−1 y ∈ H. Claim: F is 1-1 correspondence. F is onto since, for any b, Hb = F (b−1 H). F is 1-1 because F (aH) = F (bH) implies that Ha−1 = Hb−1 which by definition says that a−1 (b−1)−1 = a−1 b ∈ H. But this implies aH = bH. 1
Since F is 1-1 correspondence we have | L |=| R | .
