User interface language: English | Español

Date May 2017 Marks available 2 Reference code 17M.2.hl.TZ0.8
Level HL only Paper 2 Time zone TZ0
Command term Determine Question number 8 Adapted from N/A

Question

The set \({S_n} = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }} \ldots ,{\text{ }}n - 2,{\text{ }}n - 1\} \), where \(n\) is a prime number greater than 2, and \({ \times _n}\) denotes multiplication modulo \(n\).

Show that there are no elements \(a,{\text{ }}b \in {S_n}\) such that \(a{ \times _n}b = 0\).

[2]
a.i.

Show that, for \(a,{\text{ }}b,{\text{ }}c \in {S_n},{\text{ }}a{ \times _n}b = a{ \times _n}c \Rightarrow b = c\).

[2]
a.ii.

Show that \({G_n} = \{ {S_n},{\text{ }}{ \times _n}\} \) is a group. You may assume that \({ \times _n}\) is associative.

[4]
b.

Show that the order of the element \((n - 1)\) is 2.

[1]
c.i.

Show that the inverse of the element 2 is \(\frac{1}{2}(n + 1)\).

[2]
c.ii.

Explain why the inverse of the element 3 is \(\frac{1}{3}(n + 1)\) for some values of \(n\) but not for other values of \(n\).

[2]
c.iii.

Determine the inverse of the element 3 in \({G_{11}}\).

[1]
c.iv.

Determine the inverse of the element 3 in \({G_{31}}\).

[2]
c.v.

Markscheme

\(a{ \times _n}b = 0 \Rightarrow ab = \) a multiple of \(n\) (or vice versa)     R1

since \(n\) is prime, this can only occur if \(a = 1\) and \(b = \) multiple of \(n\) which is impossible because the multiple of \(n\) would not belong to \({S_n}\)     R1

[2 marks]

a.i.

\(a{ \times _n}b = a{ \times _n}c \Rightarrow a{ \times _n}(b - c) = 0\)     M1

suppose \(b \ne c\) and let \(b > c\) (without loss of generality)

\((b - c) \in {S_n}\) and from (i), \(a{ \times _n}(b - c) = 0\) is a contradiction     R1

therefore \(b = c\)     AG

[2 marks]

a.ii.

\({G_n}\) is associative because modular multiplication is associative     A1

\({G_n}\) is closed because the value of \(a{ \times _n}b\) always lies between 1 and \(n - 1\)     A1

the identity is 1     A1

consider \(a{ \times _n}b\) where \(b\) can take \(n - 1\) possible values. Using the result from (a)(ii), this will result in \(n - 1\) different values, one of which will be 1, which will give the inverse of \(a\)     R1

\({G_n}\) is therefore a group     AG

[4 marks]

b.

\({(n - 1)^2} = {n^2} - 2n + 1 \equiv 1(\bmod n)\)     M1

so that \((n - 1){ \times _n}(n - 1) = 1\) and \(n - 1\) has order 2     R1AG

[??? marks]

c.i.

consider \(2 \times \frac{1}{2}(n + 1) = n + 1 = 1(\bmod n)\)     A1

since \(\frac{1}{2}(n + 1)\) is an integer for al \(n\), it is the inverse of 2     R1AG

[??? marks]

c.ii.

consider \(3 \times \frac{1}{3}(n + 1) = n + 1 = 1(\bmod n)\)     M1

therefore \(\frac{1}{3}(n + 1)\) is the inverse of 3 if it is an integer but not otherwise     R1

[??? marks]

c.iii.

the inverse of 3 in \({G_{11}}\) is 4     A1

[??? marks]

c.iv.

the inverse of 3 in \({G_{31}}\) is 21     (M1)A1

[??? marks]

c.v.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
c.iii.
[N/A]
c.iv.
[N/A]
c.v.

Syllabus sections

Topic 4 - Sets, relations and groups » 4.6

View options