Date | November 2015 | Marks available | 1 | Reference code | 15N.3srg.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | State | Question number | 3 | Adapted from | N/A |
Question
The set of all permutations of the elements 1, 2, …101, 2, …10 is denoted by HH and the binary operation ∘∘ represents the composition of permutations.
The permutation p=(1 2 3 4 5 6)(7 8 9 10)p=(1 2 3 4 5 6)(7 8 9 10) generates the subgroup {G, ∘}{G, ∘} of the group {H, ∘}{H, ∘}.
Find the order of {G, ∘}{G, ∘}.
State the identity element in {G, ∘}{G, ∘}.
Find
(i) p∘pp∘p;
(ii) the inverse of p∘pp∘p.
(i) Find the maximum possible order of an element in {H, ∘}{H, ∘}.
(ii) Give an example of an element with this order.
Markscheme
the order of (G, ∘)(G, ∘) is lcm(6, 4)lcm(6, 4) (M1)
=12=12 A1
[2 marks]
(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10) A1
Note: Accept ( ) or a word description.
[1 mark]
(i) p∘p=(1 3 5)(2 4 6)(7 9)(810)p∘p=(1 3 5)(2 4 6)(7 9)(810) (M1)A1
(ii) its inverse =(1 5 3)(2 6 4)(7 9)(810)=(1 5 3)(2 6 4)(7 9)(810) A1A1
Note: Award A1 for cycles of 2, A1 for cycles of 3.
[4 marks]
(i) considering LCM of length of cycles with length 22, 33 and 55 (M1)
3030 A1
(ii) eg(1 2)(3 4 5)(6 7 8 9 10)(1 2)(3 4 5)(6 7 8 9 10) A1
Note: allow FT as long as the length of cycles adds to 1010 and their LCM is consistent with answer to part (i).
Note: Accept alternative notation for each part
[3 marks]
Total [10 marks]