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.

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]

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.