Question
The binary operation \( * \) is defined for \(a{\text{, }}b \in {\mathbb{Z}^ + }\) by
\[a * b = a + b – 2.\]
(a) Determine whether or not \( * \) is
(i) closed,
(ii) commutative,
(iii) associative.
(b) (i) Find the identity element.
(ii) Find the set of positive integers having an inverse under \( * \).
▶️Answer/Explanation
Markscheme
(a) (i) It is not closed because
\(1 * 1 = 0 \notin {\mathbb{Z}^ + }\) . R2
(ii) \(a * b = a + b – 2\)
\(b * a = b + a – 2 = a * b\) M1
It is commutative. A1
(iii) It is not associative. A1
Consider \((1 * 1) * 5\) and \(1 * (1 * 5)\) .
The first is undefined because \(1 * 1 \notin {\mathbb{Z}^ + }\) .
The second equals 3. R2
Notes: Award A1R2 for stating that non-closure implies non-associative.
Award A1R1 to candidates who show that \(a * (b * c) = (a * b) * c = a + b + c – 4\) and therefore conclude that it is associative, ignoring the non-closure.
[7 marks]
(b) (i) The identity e satisfies
\(a * e = a + e – 2 = a\) M1
\(e = 2\,\,\,\,\,({\text{and }}2 \in {\mathbb{Z}^ + })\) A1
(ii) \(a * {a^{ – 1}} = a + {a^{ – 1}} – 2 = 2\) M1
\(a + {a^{ – 1}} = 4\) A1
So the only elements having an inverse are 1, 2 and 3. A1
Note: Due to commutativity there is no need to check two sidedness of identity and inverse.
[5 marks]
Total [12 marks]
Examiners report
Almost all the candidates thought that the binary operation was associative, not realising that the non-closure prevented this from being the case. In the circumstances, however, partial credit was given to candidates who ‘proved’ associativity. Part (b) was well done by many candidates.
Question
The binary operation \( * \) is defined on \(\mathbb{R}\) as follows. For any elements a , \(b \in \mathbb{R}\)
\[a * b = a + b + 1.\]
(i) Show that \( * \) is commutative.
(ii) Find the identity element.
(iii) Find the inverse of the element a .
The binary operation \( \cdot \) is defined on \(\mathbb{R}\) as follows. For any elements a , \(b \in \mathbb{R}\)
\(a \cdot b = 3ab\) . The set S is the set of all ordered pairs \((x,{\text{ }}y)\) of real numbers and the binary operation \( \odot \) is defined on the set S as
\(({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2}) = ({x_1} * {x_2},{\text{ }}{y_1} \cdot {y_2}){\text{ }}.\)
Determine whether or not \( \odot \) is associative.
▶️Answer/Explanation
Markscheme
(i) if \( * \) is commutative \(a * b = b * a\)
since \(a + b + 1 = b + a + 1\) , \( * \) is commutative R1
(ii) let e be the identity element
\(a * e = a + e + 1 = a\) M1
\( \Rightarrow e = – 1\) A1
(iii) let a have an inverse, \({a^{ – 1}}\)
\(a * {a^{ – 1}} = a + {a^{ – 1}} + 1 = – 1\) M1
\( \Rightarrow {a^{ – 1}} = – 2 – a\) A1
[5 marks]
\(({x_1},{\text{ }}{y_1}) \odot \left( {({x_2},{\text{ }}{y_2}) \odot ({x_3},{\text{ }}{y_3})} \right) = ({x_1},{\text{ }}{y_1}) \odot ({x_2} + {x_3} + 1,{\text{ }}3{y_2}{y_3})\) M1
\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\) A1A1
\(\left( {({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2})} \right) \odot ({x_3},{\text{ }}{y_3}) = ({x_1} + {x_2} + 1,{\text{ }}3{y_1}{y_2}) \odot ({x_3},{\text{ }}{y_3})\) M1
\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\) A1
hence \( \odot \) is associative R1
[6 marks]
Examiners report
Part (a) of this question was the most accessible on the paper and was completed correctly by the majority of candidates.
Part (b) was completed by many candidates, but a significant number either did not understand what was meant by associative, confused associative with commutative, or were unable to complete the algebra.
Question
(a) Consider the set A = {1, 3, 5, 7} under the binary operation \( * \), where \( * \) denotes multiplication modulo 8.
(i) Write down the Cayley table for \(\{ A,{\text{ }} * \} \).
(ii) Show that \(\{ A,{\text{ }} * \} \) is a group.
(iii) Find all solutions to the equation \(3 * x * 7 = y\). Give your answers in the form \((x,{\text{ }}y)\).
(b) Now consider the set B = {1, 3, 5, 7, 9} under the binary operation \( \otimes \), where \( \otimes \) denotes multiplication modulo 10. Show that \(\{ B,{\text{ }} \otimes \} \) is not a group.
(c) Another set C can be formed by removing an element from B so that \(\{ C,{\text{ }} \otimes \} \) is a group.
(i) State which element has to be removed.
(ii) Determine whether or not \(\{ A,{\text{ }} * \} \) and \(\{ C,{\text{ }} \otimes \} \) are isomorphic.
▶️Answer/Explanation
Markscheme
(a) (i) 
A3
Note: Award A2 for 15 correct, A1 for 14 correct and A0 otherwise.
(ii) it is a group because:
the table shows closure A1
multiplication is associative A1
it possesses an identity 1 A1
justifying that every element has an inverse e.g. all self-inverse A1
(iii) (since \( * \) is commutative, \(5 * x = y\))
so solutions are (1, 5), (3, 7), (5, 1), (7, 3) A2
Notes: Award A1 for 3 correct and A0 otherwise.
Do not penalize extra incorrect solutions.
[9 marks]
(b) 
Note: It is not necessary to see the Cayley table.
a valid reason R2
e.g. from the Cayley table the 5 row does not give a Latin square, or 5 does not have an inverse, so it cannot be a group
[2 marks]
(c) (i) remove the 5 A1
(ii) they are not isomorphic because all elements in A are self-inverse this is not the case in C, (e.g. \(3 \otimes 3 = 9 \ne 1\)) R2
Note: Accept any valid reason.
[3 marks]
Total [14 marks]
Examiners report
Candidates are generally confident when dealing with a specific group and that was the situation again this year. Some candidates lost marks in (a)(ii) by not giving an adequate explanation for the truth of some of the group axioms, eg some wrote ‘every element has an inverse’. Since the question told the candidates that \(\{ A,{\text{ }} * )\) was a group, this had to be the case and the candidates were expected to justify their statement by noting that every element was self-inverse. Solutions to (c)(ii) were reasonably good in general, certainly better than solutions to questions involving isomorphisms set in previous years.
Question
Set \(S = \{ {x_0},{\text{ }}{x_1},{\text{ }}{x_2},{\text{ }}{x_3},{\text{ }}{x_4},{\text{ }}{x_5}\} \) and a binary operation \( \circ \) on S is defined as \({x_i} \circ {x_j} = {x_k}\), where \(i + j \equiv k(\bmod 6)\).
(a) (i) Construct the Cayley table for \(\{ S,{\text{ }} \circ \} \) and hence show that it is a group.
(ii) Show that \(\{ S,{\text{ }} \circ \} \) is cyclic.
(b) Let \(\{ G,{\text{ }} * \} \) be an Abelian group of order 6. The element \(a \in {\text{G}}\) has order 2 and the element \(b \in {\text{G}}\) has order 3.
(i) Write down the six elements of \(\{ G,{\text{ }} * \} \).
(ii) Find the order of \({\text{a}} * b\) and hence show that \(\{ G,{\text{ }} * \} \) is isomorphic to \(\{ S,{\text{ }} \circ \} \).
▶️Answer/Explanation
Markscheme
(a) (i) Cayley table for \(\{ S,{\text{ }} \circ \} \)
\(\begin{array}{*{20}{c|cccccc}}
\circ &{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\
\hline
{{x_0}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\
{{x_1}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}} \\
{{x_2}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}} \\
{{x_3}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}} \\
{{x_4}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}} \\
{{x_5}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}
\end{array}\) A4
Note: Award A4 for no errors, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.
S is closed under \( \circ \) A1
\({x_0}\) is the identity A1
\({x_0}\) and \({x_3}\) are self-inverses, A1
\({x_2}\) and \({x_4}\) are mutual inverses and so are \({x_1}\) and \({x_5}\) A1
modular addition is associative A1
hence, \(\{ S,{\text{ }} \circ \} \) is a group AG
(ii) the order of \({x_1}\) (or \({x_5}\)) is 6, hence there exists a generator, and \(\{ S,{\text{ }} \circ \} \) is a cyclic group A1R1
[11 marks]
(b) (i) e, a, b, ab A1
and \({b^2},{\text{ }}a{b^2}\) A1A1
Note: Accept \(ba\) and \({b^2}a\).
(ii) \({(ab)^2} = {b^2}\) M1A1
\({(ab)^3} = a\) A1
\({(ab)^4} = b\) A1
hence order is 6 A1
groups G and S have the same orders and both are cyclic R1
hence isomorphic AG
[9 marks]
Total [20 marks]
Examiners report
a) Most candidates had the correct Cayley table and were able to show successfully that the group axioms were satisfied. Some candidates, however, simply stated that an inverse exists for each element without stating the elements and their inverses. Most candidates were able to find a generator and hence show that the group is cyclic.
b) This part was answered less successfully by many candidates. Some failed to find all the elements. Some stated that the order of ab is 6 without showing any working.
Question
The binary operation \(\Delta\) is defined on the set \(S =\) {1, 2, 3, 4, 5} by the following Cayley table.
(a) State whether S is closed under the operation Δ and justify your answer.
(b) State whether Δ is commutative and justify your answer.
(c) State whether there is an identity element and justify your answer.
(d) Determine whether Δ is associative and justify your answer.
(e) Find the solutions of the equation \(a\Delta b = 4\Delta b\), for \(a \ne 4\).
▶️Answer/Explanation
Markscheme
(a) yes A1
because the Cayley table only contains elements of S R1
[2 marks]
(b) yes A1
because the Cayley table is symmetric R1
[2 marks]
(c) no A1
because there is no row (and column) with 1, 2, 3, 4, 5 R1
[2 marks]
(d) attempt to calculate \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) M1
counterexample: for example, \((1\Delta 2)\Delta 3 = 2\)
\(1\Delta (2\Delta 3) = 1\) A1
Δ is not associative A1
Note: Accept a correct evaluation of \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) for the M1.
[3 marks]
(e) for example, attempt to enumerate \(4\Delta b\) for b = 1, 2, 3, 4, 5 and obtain (3, 2, 1, 4, 1) (M1)
find \((a,{\text{ }}b) \in \left\{ {{\text{(2, 2), (2, 3)}}} \right\}\) for \(a \ne 4\) (or equivalent) A1A1
Note: Award M1A1A0 if extra ‘solutions’ are listed.
[3 marks]
Total [12 marks]