Question
The relation \(R\) is defined for \(x,y \in {\mathbb{Z}^ + }\) such that \(xRy\) if and only if \({3^x} \equiv {3^y}(\bmod 10)\) .
(i) Show that \(R\) is an equivalence relation.
(ii) Identify all the equivalence classes.
Let \(S\) denote the set \(\left\{ {x\left| {x = a + b\sqrt 3 ,a,b \in \mathbb{Q},{a^2} + {b^2} \ne 0} \right.} \right\}\) .
(i) Prove that \(S\) is a group under multiplication.
(ii) Give a reason why \(S\) would not be a group if the conditions on \({a,b}\) were changed to \({a,b \in \mathbb{R},{a^2} + {b^2} \ne 0}\) .
Answer/Explanation
Markscheme
(i) \({3^x} \equiv {3^x}(\bmod 10) \Rightarrow xRx\) so R is reflexive. R1
\(xRy \Rightarrow {3^x} \equiv {3^y}(\bmod 10) \Rightarrow {3^y} \equiv {3^x}(\bmod 10) \Rightarrow yRx\)
so \(R\) is symmetric. R2
\(xRy\) and \(yRz \Rightarrow {3^x} – {3^y} = 10M\) and \({3^y} – {3^z} = 10N\)
Adding \({3^x} – {3^z} = 10(M + N) \Rightarrow {3^x} \equiv {3^z}(\bmod 10)\) hence transitive R2
(ii) Consider \({3^1} = 3,{3^2} = 9,{3^3} = 27,{3^4} = 81,{3^5} = 243\) , etc. (M2)
It is evident from this sequence that there are 4 equivalence classes,
\(1\), \(5\), \(9\), … A1
\(2\), \(6\), \(10\), … A1
\(3\), \(7\), \(11\), … A1
\(4\), \(8\), \(12\), … A1
[11 marks]
(i) Consider \(a + b\sqrt 3 \) \(c + d\sqrt 3 \) \( = (ac + 3bd) + (bc + ad)\sqrt 3 \) M1A1
This establishes closure since products of rational numbers are rational. R1
Since if \(a\) and \(b\) are not both zero and \(c\) and \(d\) are not both zero, it follows that \(ac + 3bd\) and \(bc + ad\) are not both zero. R1
The identity is \(1( \in S)\) . R1
Consider \(a + b{\sqrt 3 ^{ – 1}} = \frac{1}{{a + b\sqrt 3 }}\) M1A1
\( = \frac{1}{{a + b\sqrt 3 }} \times \frac{{a – b\sqrt 3 }}{{a – b\sqrt 3 }}\) A1
\( = \frac{a}{{({a^2} – 3{b^2})}} \times \frac{b}{{({a^2} – 3{b^2})}}\sqrt 3 \) A1
This inverse \( \in S\) because \({({a^2} – 3{b^2})}\) cannot equal zero since \(a\) and \(b\) cannot both be zero R1
and \(({a^2} – 3{b^2}) = 0\) would require \(\frac{a}{b} = \pm \sqrt 3 \) which is impossible because a rational number cannot equal \(\sqrt 3 \) . R2
Finally, multiplication of numbers is associative. R1
(ii) If \(a\) and \(b\) are both real numbers, \(a + b\sqrt 3 \) would have no inverse if \({a^2} = 3{b^2}\) . R2
[15 marks]
Question
A group has exactly three elements, the identity element \(e\) , \(h\) and \(k\) . Given the operation is denoted by \( \otimes \) , show that
(i) Show that \({\mathbb{Z}_4}\) (the set of integers modulo 4) together with the operation \({ + _4}\) (addition modulo 4) form a group \(G\) . You may assume associativity.
(ii) Show that \(G\) is cyclic.
Using Cayley tables or otherwise, show that \(G\) and \(H = \left( {\left\{ {1,2,3,\left. 4 \right\},{ \times _5}} \right.} \right)\) are isomorphic where \({{ \times _5}}\) is multiplication modulo 5. State clearly all the possible bijections.
the group is cyclic.
the group is cyclic.
Answer/Explanation
Markscheme
(i)
A2
Note: Award A1 for table if exactly one error and A0 if more than one error.
all elements belong to \({\mathbb{Z}_4}\) so it is closed A1
\(0\) is the identity element A1
\(2\) is self inverse A1
\(1\) and \(3\) are an inverse pair A1
hence every element has an inverse
hence \(\left\{ {{\mathbb{Z}_4},{ + _4}} \right\}\) form a group \(G\) AG
(ii) \(1{ + _4}1 \equiv 2(\bmod 4)\)
\(1{ + _4}1{ + _4}1 \equiv 3(\bmod 4)\)
\(1{ + _4}1{ + _4}1{ + _4}1 \equiv 0(\bmod 4)\) M1A1
hence \(1\) is a generator R1
therefore \(G\) is cyclic AG
(\(3\) is also a generator)
[9 marks]
A1A1
EITHER
for the group \(\left( {\left\{ {1,2,3,\left. 4 \right\},{ \times _5}} \right.} \right)\)
\(1\) is the identity and \(4\) is self inverse A1
\(2\) and \(3\) are an inverse pair A1
OR
for \(G\), for \(H\),
0 has order 1 1 has order 1
1 has order 4 2 has order 4
2 has order 2 3 has order 4
3 has order 4 4 has order 2 A1A1
THEN
hence there is a bijection R1
\(h(1) \to 0\) , \(h(2) \to 1\) , \(h(3) \to 3\) , \(h(4) \to 2\) A1
the groups are isomorphic AG
\(k(1) \to 0\) , \(k(2) \to 3\) , \(k(3) \to 1\) , \(k(4) \to 2\) A1
is also a bijection
[7 marks]
if cyclic then the group is {\(e\), \(h\), \({h^2}\)} R1
\({h^2} = e\) or \(h\) or \(k\) M1
\({h^2} = e \Rightarrow h \otimes h = h \otimes k\)
\( \Rightarrow h = k\)
but \(h \ne k\) so \({h^2} \ne e\) A1
\({h^2} = h \Rightarrow h \otimes h = h \otimes e \Rightarrow h = e\)
but \(h \ne e\) so \({h^2} \ne h\)
so \({h^2} = k\) A1
also \({h^3} = h \otimes k = e\)
hence the group is cyclic AG
Note: An alternative proof is possible based on order of elements and Lagrange.
[5 marks]
if cyclic then the group is \(\left\{ {e,h,\left. {{h^2}} \right\}} \right.\) R1
\({h^2} = e\) or \(h\) or \(k\) M1
\({h^2} = e \Rightarrow h \otimes h = h \otimes k\)
\( \Rightarrow h = k\)
but \(h \ne k\) so \({h^2} \ne e\) A1
\({h^2} = h \Rightarrow h \otimes h = h \otimes e \Rightarrow h = e\)
but \(h \ne e\) so \({h^2} \ne h\)
so \({h^2} = k\) A1
also \({h^3} = h \otimes k = e\) A1
hence the group is cyclic AG
Note: An alternative proof is possible based on order of elements and Lagrange.
[5 marks]
Question
The binary operator \( * \) is defined for a , \(b \in \mathbb{R}\) by \(a * b = a + b – ab\) .
(i) Show that \( * \) is associative.
(ii) Find the identity element.
(iii) Find the inverse of \(a \in \mathbb{R}\) , showing that the inverse exists for all values of \(a\) except one value which should be identified.
(iv) Solve the equation \(x * x = 1\) .
The domain of \( * \) is now reduced to \(S = \left\{ {0,2,3,4,5,\left. 6 \right\}} \right.\) and the arithmetic is carried out modulo \(7\).
(i) Copy and complete the following Cayley table for \(\left\{ {S,\left. * \right\}} \right.\) .
(ii) Show that \(\left\{ {S,\left. * \right\}} \right.\) is a group.
(iii) Determine the order of each element in S and state, with a reason, whether or not \(\left\{ {S,\left. * \right\}} \right.\) is cyclic.
(iv) Determine all the proper subgroups of \(\left\{ {S,\left. * \right\}} \right.\) and explain how your results illustrate Lagrange’s theorem.
(v) Solve the equation \(2 * x * x = 5\) .
Answer/Explanation
Markscheme
(i) \(a * (b * c) = a * (b + c – bc)\) M1
\( = a + b + c – bc – a(b + c – bc)\) A1
\( = a + b + c – bc – ca – ab + abc\) A1
\((a * b) * c = (a + b – ab) * c\) M1
\( = a + b – ab + c – (a + b – ab)c\) A1
\( = a + b + c – bc – ca – ab + abc\) , hence associative AG
(ii) let \(e\) be the identity element, so that \(a * e = a\) (M1)
then,
\(a + e – ae = a\) A1
\(e(1 – a) = 0\)
\(e = 0\) A1
(iii) let \({a^{ – 1}}\) be the inverse of \(a\), so that \(a * {a^{ – 1}} = 0\) (M1)
then,
\(a + {a^{ – 1}} – a{a^{ – 1}} = 0\) A1
\({a^{ – 1}} = \frac{a}{{a – 1}}\) A1
this gives an inverse for all elements except 1 which has no inverse R1
(iv) \(2x – {x^2} = 1\) M1
\({(x – 1)^2} = 0\) (A1)
\(x = 1\) A1
[15 marks]
(i)
A3
Note: Award A3 for correct table, A2 for one error, A1 for two errors and A0 for more than two errors.
(ii) there are no new elements in the table so it is closed A1
there is an identity element, \(0\) A1
every row (column) has a \(0\) so every element has an inverse A1
associativity has been proved earlier A1
therefore \(\left\{ {S,\left. * \right\}} \right.\) is a group AG
(iii)
A3
Note: Award A3 for correct table, A2 for one error, A1 for two errors and A0 for more than two errors.
it is cyclic because there are elements of order \(6\) R1
(iv) the proper subgroups are \(\left\{ {0,\left. 2 \right\}} \right.\) , \(\left\{ {0,\left. {4,6} \right\}} \right.\) A1A1
the orders of the subgroups (\(2\), \(3\)) are factors of the order of the group (6) A1
(v) recognizing \(x * x = 4\) (M1)
\(x = 3\) , \(6\) A1A1
[17 marks]
Question
(i) Draw the Cayley table for the set \(S = \left\{ {0,1,2,3,4,\left. 5 \right\}} \right.\) under addition modulo six \(({ + _6})\) and hence show that \(\left\{ {S, + \left. {_6} \right\}} \right.\) is a group.
(ii) Show that the group is cyclic and write down its generators.
(iii) Find the subgroup of \(\left\{ {S, + \left. {_6} \right\}} \right.\) that contains exactly three elements.
Prove that a cyclic group with exactly one generator cannot have more than two elements.
\(H\) is a group and the function \(\Phi :H \to H\) is defined by \(\Phi (a) = {a^{ – 1}}\) , where \({a^{ – 1}}\) is the inverse of a under the group operation. Show that \(\Phi \) is an isomorphism if and only if H is Abelian.
Answer/Explanation
Markscheme
(i)
the table is closed A1
the identity is \(0\) A1
\(0\) is in every row and column once so each element has a unique inverse A1
addition is associative A1
therefore \(\left\{ {S, + \left. {_6} \right\}} \right.\) is a group R1
(ii) \(1 + 1 + 1 + 1 + 1 + 1 = 0\) M1
\(1 + 1 + 1 + 1 + 1 = 5\)
\(1 + 1 + 1 + 1 = 4\)
\(1 + 1 + 1 = 3\)
\(1 + 1 = 2\)
so \(1\) is a generator of \(\left\{ {S, + \left. {_6} \right\}} \right.\) and the group is cyclic A1
(since \(5\) is the additive inverse of \(1\)) \(5\) is also a generator A1
(iii) \(\left\{ {0,2,\left. 4 \right\}} \right.\) A1
[11 marks]
if \(a\) is a generator of group \((G, * )\) then so is \({a^{ – 1}}\) A1
if \((G, * )\) has exactly one generator \(a\) then \(a = {a^{ – 1}}\) A1
so \({a^2} = e\) and \(G = \left\{ {e,\left. a \right\}} \right.\) \(\left\{ {\left. e \right\}} \right.\) A1R1
so cyclic group with exactly one generator cannot have more than two elements AG
[4 marks]
every element of a group has a unique inverse so \(\Phi \) is a bijection A1
\(\Phi (ab) = {(ab)^{ – 1}} = {b^{ – 1}}{a^{ – 1}}\) M1A1
if \(H\) is Abelian then it follows that
\({b^{ – 1}}{a^{ – 1}} = {a^{ – 1}}{b^{ – 1}} = \Phi (a)\Phi (b)\) A1
so \(\Phi \) is an isomorphism R1
if \(\Phi \) is an isomorphism, then M1
for all \(a,b \in H\) , \(\Phi (ab) = \Phi (a)\Phi (b)\) M1
\({(ab)^{ – 1}} = {a^{ – 1}}{b^{ – 1}}\)
\( \Rightarrow {b^{ – 1}}{a^{ – 1}} = {a^{ – 1}}{b^{ – 1}}\) A1
so \(H\) is Abelian R1
[9 marks]
Question
The set \(S\) consists of real numbers r of the form \(r = a + b\sqrt 2 \) , where \(a,b \in \mathbb{Z}\) .
The relation \(R\) is defined on \(S\) by \({r_1}R{r_2}\) if and only if \({a_1} \equiv {a_2}\) (mod2) and \({b_1} \equiv {b_2}\) (mod3), where \({r_1} = {a_1} + {b_1}\sqrt 2 \) and \({r_2} = {a_2} + {b_2}\sqrt 2 \) .
Show that \(R\) is an equivalence relation.
Show, by giving a counter-example, that the statement \({r_1}R{r_2} \Rightarrow r_1^2Rr_2^2\) is false.
Determine
(i) the equivalence class \(E\) containing \(1 + \sqrt 2 \) ;
(ii) the equivalence class \(F\) containing \(1 – \sqrt 2 \) .
Show that
(i) \({(1 + \sqrt 2 )^3} \in F\) ;
(ii) \({(1 + \sqrt 2 )^6} \in E\) .
Determine whether the set \(E\) forms a group under
(i) the operation of addition;
(ii) the operation of multiplication.
Answer/Explanation
Markscheme
reflexive: if \({r_{}} = {a_{}} + {b_{}}\sqrt 2 \in S\) then \(a \equiv a(\bmod 2)\) and \(b \equiv b(\bmod 3)\)
\(( \Rightarrow rRr)\) A1
symmetric: if \({r_1}R{r_2}\) then \({a_1} \equiv {a_2}(\bmod 2)\) and \({b_1} \equiv {b_2}(\bmod 3)\) , and M1
\({a_2} \equiv {a_1}(\bmod 2)\) and \({b_2} \equiv {b_1}(\bmod 3)\) , (so that \({r_2}R{r_1}\) ) A1
transitive: if \({r_1}R{r_2}\) and \({r_2}R{r_3}\) then
\(2|{a_1} – {a_2}\) and \(2|{a_2} – {a_3}\) M1
\( \Rightarrow 2|{a_1} – {a_2} + {a_2} – {a_3} \Rightarrow 2|{a_1} – {a_3}\) M1A1
\(3|{b_1} – {b_2}\) and \(3|{b_2} – {b_3}\)
\( \Rightarrow 3|{b_1} – {b_2} + {b_2} – {b_3} \Rightarrow 3|{b_1} – {b_3}( \Rightarrow {r_1}R{r_3})\) A1AG
[7 marks]
consider, for example, \({r_1} = 1 + \sqrt 2 \) , \({r_2} = 3 + \sqrt 2 \) \(({r_1}R{r_2})\) M1
Note: Only award M1 if the two numbers are related and neither \(a\) nor \(b = 0\) .
\(r_1^2 = 3 + 2\sqrt 2 \) , \(r_2^2 = 11 + 6\sqrt 2 \) A1
the squares are not equivalent because \(2 \ne 6(\bmod 3)\) A1
[3 marks]
(i) \(E = \left\{ {2k + 1 + (3m + 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) A1A1
(ii) \(F = \left\{ {2k + 1 + (3m – 1)\sqrt 2 } :k,m \in \mathbb{Z} \right\}\) A1
[3 marks]
(i) \({(1 + \sqrt 2 )^3} = 7 + 5\sqrt 2 \) A1
\( = 2 \times 3 + 1 + (3 \times 2 – 1)\sqrt 2 \in F\) R1AG
(ii) \({(1 + \sqrt 2 )^6} = 99 + 70\sqrt 2 \) A1
\( = 2 \times 49 + 1 + (3 \times 23 + 1)\sqrt 2 \in E\) R1AG
[4 marks]
(i) \(E\) is not a group under addition A1
any valid reason eg \(0 \notin E\) R1
(ii) \(E\) is not a group under multiplication A1
any valid reason eg \(1 \notin E\) R1
[4 marks]
Question
The set \(S\) contains the eighth roots of unity given by \(\left\{ {{\text{cis}}\left( {\frac{{n\pi }}{4}} \right),{\text{ }}n \in \mathbb{N},{\text{ }}0 \leqslant n \leqslant 7} \right\}\).
(i) Show that \(\{ S,{\text{ }} \times \} \) is a group where \( \times \) denotes multiplication of complex numbers.
(ii) Giving a reason, state whether or not \(\{ S,{\text{ }} \times \} \) is cyclic.
Answer/Explanation
Markscheme
(i) closure: let \({a_1} = {\text{cis}}\left( {\frac{{{n_1}\pi }}{4}} \right)\) and \({a_2} = {\text{cis}}\left( {\frac{{{n_2}\pi }}{4}} \right) \in S\) M1
then \({a_1} \times {a_2} = {\text{cis}}\left( {\frac{{({n_1} + {n_2})\pi }}{4}} \right)\) (which \( \in S\) because the addition is carried out modulo 8) A1
identity: the identity is 1 (and corresponds to \(n = 0\)) A1
inverse: the inverse of \({\text{cis}}\left( {\frac{{n\pi }}{4}} \right)\) is \({\text{cis}}\left( {\frac{{(8 – n)\pi }}{4}} \right) \in S\) A1
associatively: multiplication of complex numbers is associative A1
the four group axioms are satisfied so \(S\) is a group AG
(ii) \(S\) is cyclic A1
because \({\text{cis}}\left( {\frac{\pi }{4}} \right)\), for example, is a generator R1
[7 marks]
Question
The binary operation multiplication modulo \(9\), denoted by \({ \times _9}\) , is defined on the set \(S = \left\{ {1,2,3,4,5,6,7,8} \right\}\) .
Copy and complete the following Cayley table.
Show that \(\left\{ {S,{ \times _9}} \right\}\) is not a group.
Prove that a group \(\left\{ {G,{ \times _9}} \right\}\) can be formed by removing two elements from the set \(S\) .
(i) Find the order of all the elements of \(G\) .
(ii) Write down all the proper subgroups of \(\left\{ {G,{ \times _9}} \right\}\) .
(iii) Determine the coset containing the element \(5\) for each of the subgroups in part (ii).
Solve the equation \(4{ \times _9}x{ \times _9}x = 1\) .
Answer/Explanation
Markscheme
A3
Note: Award A2 if one error, A1 if two errors and A0 if three or more errors.
[3 marks]
any valid reason, R1
e.g. not closed
\(3\) or \(6\) has no inverse,
it is not a Latin square
[1 mark]
remove \(3\) and \(6\) A1
for the remaining elements,
the table is closed R1
associative because multiplication is associative R1
the identity is \(1\) A1
every element has an inverse, (\(2\), \(5\)) and (\(4\), \(7\)) are inverse pairs and \(8\) (and \(1\)) are self-inverse A1
thus it is a group AG
[5 marks]
(i) the orders are
A3
Note: Award A2 if one error, A1 if two errors and A0 if three or more errors.
(ii) the proper subgroups are
\(\left\{ {1,8} \right\}\) A1
\(\left\{ {1,4,7} \right\}\) A1
Note: Do not penalize inclusion of \(\left\{ 1 \right\}\) .
(iii) the cosets are \(\left\{ {5,4} \right\}\) (M1)A1
\(\left\{ {5,2,8} \right\}\) A1
[8 marks]
\(x{ \times _9}x = 7\) (A1)
\(x = 4,5\) A1A1
[3 marks]