| Date | November 2015 | Marks available | 4 | Reference code | 15N.3srg.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The set of all permutations of the elements \(1,{\text{ }}2,{\text{ }} \ldots 10\) is denoted by \(H\) and the binary operation \( \circ \) represents the composition of permutations.
The permutation \(p = (1{\text{ }}2{\text{ }}3{\text{ }}4{\text{ }}5{\text{ }}6)(7{\text{ }}8{\text{ }}9{\text{ }}10)\) generates the subgroup \(\{ G,{\text{ }} \circ \} \) of the group \(\{ H,{\text{ }} \circ \} \).
Find the order of \(\{ G,{\text{ }} \circ \} \).
State the identity element in \(\{ G,{\text{ }} \circ \} \).
Find
(i) \(p \circ p\);
(ii) the inverse of \(p \circ p\).
(i) Find the maximum possible order of an element in \(\{ H,{\text{ }} \circ \} \).
(ii) Give an example of an element with this order.
Markscheme
the order of \((G,{\text{ }} \circ )\) is \({\text{lcm}}(6,{\text{ }}4)\) (M1)
\( = 12\) A1
[2 marks]
\(\left( 1 \right){\rm{ }}\left( 2 \right){\rm{ }}\left( 3 \right){\rm{ }}\left( 4 \right){\rm{ }}\left( 5 \right){\rm{ }}\left( 6 \right){\rm{ }}\left( 7 \right){\rm{ }}\left( 8 \right){\rm{ }}\left( 9 \right){\rm{ }}\left( {10} \right)\) A1
Note: Accept ( ) or a word description.
[1 mark]
(i) \(p \circ p = (1{\text{ }}3{\text{ }}5)(2{\text{ }}4{\text{ }}6)(7{\text{ }}9)(810)\) (M1)A1
(ii) its inverse \( = (1{\text{ }}5{\text{ }}3)(2{\text{ }}6{\text{ }}4)(7{\text{ }}9)(810)\) A1A1
Note: Award A1 for cycles of 2, A1 for cycles of 3.
[4 marks]
(i) considering LCM of length of cycles with length \(2\), \(3\) and \(5\) (M1)
\(30\) A1
(ii) eg\(\;\;\;(1{\text{ }}2)(3{\text{ }}4{\text{ }}5)(6{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}10)\) A1
Note: allow FT as long as the length of cycles adds to \(10\) and their LCM is consistent with answer to part (i).
Note: Accept alternative notation for each part
[3 marks]
Total [10 marks]
