| Date | May 2015 | Marks available | 5 | Reference code | 15M.3srg.hl.TZ0.2 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Deduce and Find | Question number | 2 | Adapted from | N/A |
Question
The binary operation \( * \) is defined for \(x,{\text{ }}y \in S = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6\} \) by
\[x * y = ({x^3}y - xy)\bmod 7.\]
Find the element \(e\) such that \(e * y = y\), for all \(y \in S\).
(i) Find the least solution of \(x * x = e\).
(ii) Deduce that \((S,{\text{ }} * )\) is not a group.
Determine whether or not \(e\) is an identity element.
Markscheme
attempt to solve \({e^3}y - ey \equiv y\bmod 7\) (M1)
the only solution is \(e = 5\) A1
[2 marks]
(i) attempt to solve \({x^4} - {x^2} \equiv 5\bmod 7\) (M1)
least solution is \(x = 2\) A1
(ii) suppose \((S,{\text{ }} * )\) is a group with order 7 A1
\(2\) has order \(2\) A1
since \(2\) does not divide \(7\), Lagrange’s Theorem is contradicted R1
hence, \((S,{\text{ }} * )\) is not a group AG
[5 marks]
(\(5\) is a left-identity), so need to test if it is a right-identity:
ie, is \(y * 5 = y\)? M1
\(1 * 5 = 0 \ne 1\) A1
so \(5\) is not an identity A1
[3 marks]
Total [10 marks]
Examiners report
Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange's theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.
Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange's theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.
Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange's theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.
