| Date | May 2017 | Marks available | 2 | Reference code | 17M.2.hl.TZ0.8 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Explain | 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 \(a{ \times _n}b = 0\).
Show that, for \(a,{\text{ }}b,{\text{ }}c \in {S_n},{\text{ }}a{ \times _n}b = a{ \times _n}c \Rightarrow b = c\).
Show that \({G_n} = \{ {S_n},{\text{ }}{ \times _n}\} \) is a group. You may assume that \({ \times _n}\) is associative.
Show that the order of the element \((n - 1)\) is 2.
Show that the inverse of the element 2 is \(\frac{1}{2}(n + 1)\).
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 \(n\).
Determine the inverse of the element 3 in \({G_{11}}\).
Determine the inverse of the element 3 in \({G_{31}}\).
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]
\(a{ \times _n}b = a{ \times _n}c \Rightarrow a{ \times _n}(b - c) = 0\) 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]
\({G_n}\) 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]
\({(n - 1)^2} = {n^2} - 2n + 1 \equiv 1(\bmod n)\) M1
so that \((n - 1){ \times _n}(n - 1) = 1\) and \(n - 1\) has order 2 R1AG
[??? marks]
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]
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]
the inverse of 3 in \({G_{11}}\) is 4 A1
[??? marks]
the inverse of 3 in \({G_{31}}\) is 21 (M1)A1
[??? marks]
