Question
A binary operation is defined on {−1, 0, 1} by
\[A \odot B = \left\{ {\begin{array}{*{20}{c}}
{ – 1,}&{{\text{if }}\left| A \right| < \left| B \right|} \\
{0,}&{{\text{if }}\left| A \right| = \left| B \right|} \\
{1,}&{{\text{if }}\left| A \right| > \left| B \right|{\text{.}}}
\end{array}} \right.\]
(a) Construct the Cayley table for this operation.
(b) Giving reasons, determine whether the operation is
(i) closed;
(ii) commutative;
(iii) associative.
▶️Answer/Explanation
Markscheme
(a) the Cayley table is
\(\begin{gathered}
\begin{array}{*{20}{c}}
{}&{ – 1}&0&1
\end{array} \\
\begin{array}{*{20}{c}}
{ – 1} \\
0 \\
1
\end{array}\left( {\begin{array}{*{20}{c}}
0&1&0 \\
{ – 1}&0&{ – 1} \\
0&1&0
\end{array}} \right) \\
\end{gathered} \) M1A2
Notes: Award M1 for setting up a Cayley table with labels.
Deduct A1 for each error or omission.
[3 marks]
(b) (i) closed A1
because all entries in table belong to {–1, 0, 1} R1
(ii) not commutative A1
because the Cayley table is not symmetric, or counter-example given R1
(iii) not associative A1
for example because M1
\(0 \odot ( – 1 \odot 0) = 0 \odot 1 = – 1\)
but
\((0 \odot – 1) \odot 0 = – 1 \odot 0 = 1\) A1
or alternative counter-example
[7 marks]
Total [10 marks]
Examiners report
This question was generally well done, with the exception of part(b)(iii), showing that the operation is non-associative.
Question
(a) Show that {1, −1, i, −i} forms a group of complex numbers G under multiplication.
(b) Consider \(S = \{ e,{\text{ }}a,{\text{ }}b,{\text{ }}a * b\} \) under an associative operation \( * \) where e is the identity element. If \(a * a = b * b = e\) and \(a * b = b * a\) , show that
(i) \(a * b * a = b\) ,
(ii) \(a * b * a * b = e\) .
(c) (i) Write down the Cayley table for \(H = \{ S{\text{ , }} * \} \).
(ii) Show that H is a group.
(iii) Show that H is an Abelian group.
(d) For the above groups, G and H , show that one is cyclic and write down why the other is not. Write down all the generators of the cyclic group.
(e) Give a reason why G and H are not isomorphic.
▶️Answer/Explanation
Markscheme
(a)
see the Cayley table, (since there are no new elements) the set is closed A1
1 is the identity element A1
1 and –1 are self inverses and i and -i form an inverse pair, hence every element has an inverse A1
multiplication is associative A1
hence {1, –1, i, –i} form a group G under the operation of multiplication AG
[4 marks]
(b) (i) aba = aab
= eb A1
= b AG
(ii) abab = aabb
= ee A1
= e AG
[2 marks]
(c) (i)
A2
Note: Award A1 for 1 or 2 errors, A0 for more than 2.
(ii) see the Cayley table, (since there are no new elements) the set is closed A1
H has an identity element e A1
all elements are self inverses, hence every element has an inverse A1
the operation is associative as stated in the question
hence {e , a , b , ab} forms a group G under the operation \( * \) AG
(iii) since there is symmetry across the leading diagonal of the group table, the group is Abelian A1
[6 marks]
(d) consider the element i from the group G (M1)
\({{\text{i}}^2} = – 1\)
\({{\text{i}}^3} = – {\text{i}}\)
\({{\text{i}}^4} = 1\)
thus i is a generator for G and hence G is a cyclic group A1
–i is the other generator for G A1
for the group H there is no generator as all the elements are self inverses R1
[4 marks]
(e) since one group is cyclic and the other group is not, they are not isomorphic R1
[1 mark]
Total [17 marks]
Examiners report
Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. A number of candidates did not understand the term “Abelian”. Many candidates understood the conditions for a group to be cyclic. Many candidates did not realise that the answer to part (e) was actually found in part (d), hence the reason for this part only being worth 1 mark. Overall, a number of fully correct solutions to this question were seen.
Question
The binary operation \( * \) is defined on the set S = {0, 1, 2, 3} by
\[a * b = a + 2b + ab(\bmod 4){\text{ .}}\]
(a) (i) Construct the Cayley table.
(ii) Write down, with a reason, whether or not your table is a Latin square.
(b) (i) Write down, with a reason, whether or not \( * \) is commutative.
(ii) Determine whether or not \( * \) is associative, justifying your answer.
(c) Find all solutions to the equation \(x * 1 = 2 * x\) , for \(x \in S\) .
▶️Answer/Explanation
Markscheme
(a) (i)
A3
Note: Award A3 for no errors, A2 for one error, A1 for two errors and A0 for three or more errors.
(ii) it is not a Latin square because some rows/columns contain the same digit more than once A1
[4 marks]
(b) (i) EITHER
it is not commutative because the table is not symmetric about the leading diagonal R2
OR
it is not commutative because \(a + 2b + ab \ne 2a + b + ab\) in general R2
Note: Accept a counter example e.g. \(1 * 2 = 3\) whereas \(2 * 1 = 2\) .
(ii) EITHER
for example \((0 * 1) * 1 = 2 * 1 = 2\) M1
and \(0 * (1 * 1) = 0 * 0 = 0\) A1
so \( * \) is not associative A1
OR
associative if and only if \(a * (b * c) = (a * b) * c\) M1
which gives
\(a + 2b + 4c + 2bc + ab + 2ac + abc = a + 2b + ab + 2c + ac + 2bc + abc\) A1
so \( * \) is not associative as \(2ac \ne 2c + ac\) , in general A1
[5 marks]
(c) x = 0 is a solution A2
x = 2 is a solution A2
[4 marks]
Total [13 marks]
Examiners report
This question was generally well answered.
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
The binary operator multiplication modulo 14, denoted by \( * \), is defined on the set S = {2, 4, 6, 8, 10, 12}.
Copy and complete the following operation table.
(i) Show that {S , \( * \)} is a group.
(ii) Find the order of each element of {S , \( * \)}.
(iii) Hence show that {S , \( * \)} is cyclic and find all the generators.[11]
The set T is defined by \(\{ x * x:x \in S\} \). Show that {T , \( * \)} is a subgroup of {S , \( * \)}.[3]
▶️Answer/Explanation
Markscheme
A4
Note: Award A4 for all correct, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.
[4 marks]
(i) closure: there are no new elements in the table A1
identity: 8 is the identity element A1
inverse: every element has an inverse because there is an 8 in every row and column A1
associativity: (modulo) multiplication is associative A1
therefore {S , \( * \)} is a group AG
(ii) the orders of the elements are as follows
A4
Note: Award A4 for all correct, A3 for one error, A2 for two errors, A1 for three errors and A0 for four or more errors.
(iii) EITHER
the group is cyclic because there are elements of order 6 R1
OR
the group is cyclic because there are generators R1
THEN
10 and 12 are the generators A1A1
[11 marks]
looking at the Cayley table, we see that
T = {2, 4, 8} A1
this is a subgroup because it contains the identity element 8, no new elements are formed and 2 and 4 form an inverse pair R2
Note: Award R1 for any two conditions
[3 marks]
Examiners report
Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.
Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.
Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of T. This approach was invariably unsuccessful.
Question
Associativity and commutativity are two of the five conditions for a set S with the binary operation \( * \) to be an Abelian group; state the other three conditions.
The Cayley table for the binary operation \( \odot \) defined on the set T = {p, q, r, s, t} is given below.
(i) Show that exactly three of the conditions for {T , \( \odot \)} to be an Abelian group are satisfied, but that neither associativity nor commutativity are satisfied.
(ii) Find the proper subsets of T that are groups of order 2, and comment on your result in the context of Lagrange’s theorem.
(iii) Find the solutions of the equation \((p \odot x) \odot x = x \odot p\) .
▶️Answer/Explanation
Markscheme
closure, identity, inverse A2
Note: Award A1 for two correct properties, A0 otherwise.
[2 marks]
(i) closure: there are no extra elements in the table R1
identity: s is a (left and right) identity R1
inverses: all elements are self-inverse R1
commutative: no, because the table is not symmetrical about the leading diagonal, or by counterexample R1
associativity: for example, \((pq)t = rt = p\) M1A1
not associative because \(p(qt) = pr = t \ne p\) R1
Note: Award M1A1 for 1 complete example whether or not it shows non-associativity.
(ii) \(\{ s,\,p\} ,{\text{ }}\{ s,\,q\} ,{\text{ }}\{ s,\,r\} ,{\text{ }}\{ s,\,t\} \) A2
Note: Award A1 for 2 or 3 correct sets.
as 2 does not divide 5, Lagrange’s theorem would have been contradicted if T had been a group R1
(iii) any attempt at trying values (M1)
the solutions are q, r, s and t A1A1A1A1
Note: Deduct A1 if p is included.
[15 marks]
Examiners report
This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.
This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.
Question
Let \(A = \left\{ {a,{\text{ }}b} \right\}\).
Let the set of all these subsets be denoted by \(P(A)\) . The binary operation symmetric difference, \(\Delta\) , is defined on \(P(A)\) by \(X\Delta Y = (X\backslash Y) \cup (Y\backslash X)\) where \(X\) , \(Y \in P(A)\).
Let \({\mathbb{Z}_4} = \left\{ {0,{\text{ }}1,{\text{ }}2,{\text{ }}3} \right\}\) and \({ + _4}\) denote addition modulo \(4\).
Let \(S\) be any non-empty set. Let \(P(S)\) be the set of all subsets of \(S\) . For the following parts, you are allowed to assume that \(\Delta\), \( \cup \) and \( \cap \) are associative.
Write down all four subsets of A .[1]
Construct the Cayley table for \(P(A)\) under \(\Delta \) .[3]
Prove that \(\left\{ {P(A),{\text{ }}\Delta } \right\}\) is a group. You are allowed to assume that \(\Delta \) is associative.[3]
Is \(\{ P(A){\text{, }}\Delta \} \) isomorphic to \(\{ {\mathbb{Z}_4},{\text{ }}{ + _4}\} \) ? Justify your answer.[2]
(i) State the identity element for \(\{ P(S){\text{, }}\Delta \} \).
(ii) Write down \({X^{ – 1}}\) for \(X \in P(S)\) .
(iii) Hence prove that \(\{ P(S){\text{, }}\Delta \} \) is a group.[4]
Explain why \(\{ P(S){\text{, }} \cup \} \) is not a group.[1]
Explain why \(\{ P(S){\text{, }} \cap \} \) is not a group.[1]
▶️Answer/Explanation
Markscheme
\(\emptyset {\text{, \{ a\} , \{ b\} , \{ a, b\} }}\) A1
[1 mark]
A3
Note: Award A2 for one error, A1 for two errors, A0 for more than two errors.
[3 marks]
closure is seen from the table above A1
\(\emptyset \) is the identity A1
each element is self-inverse A1
Note: Showing each element has an inverse is sufficient.
associativity is assumed so we have a group AG
[3 marks]
not isomorphic as in the above group all elements are self-inverse whereas in \(({\mathbb{Z}_4},{\text{ }}{ + _4})\) there is an element of order 4 (e.g. 1) R2
[2 marks]
(i) \(\emptyset \) is the identity A1
(ii) \({X^{ – 1}} = X\) A1
(iii) if X and Y are subsets of S then \(X\Delta Y\) (the set of elements that belong to X or Y but not both) is also a subset of S, hence closure is proved R1
\(\{ P(S){\text{, }}\Delta \} \) is a group because it is closed, has an identity, all elements have inverses (and \(\Delta \) is associative) R1AG
[4 marks]
not a group because although the identity is \(\emptyset {\text{, if }}X \ne \emptyset \) it is impossible to find a set Y such that \(X \cup Y = \emptyset \), so there are elements without an inverse R1AG
[1 mark]
not a group because although the identity is S, if \(X \ne S\) is impossible to find a set Y such that \(X \cap Y = S\), so there are elements without an inverse R1AG
[1 mark]
Examiners report
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).
A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).