Question
The relation \({R_1}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_1}b\) if and only if \(n\left| {({a^2} – {b^2})} \right.\) where \(n\) is a fixed positive integer.
(i) Show that \({R_1}\) is an equivalence relation.
(ii) Determine the equivalence classes when \(n = 8\) .
Consider the group \(\left\{ {G, * } \right\}\) and let \(H\) be a subset of \(G\) defined by
\(H = \left\{ {x \in G} \right.\) such that \(x * a = a * x\) for all \(a \in \left. G \right\}\) .
Show that \(\left\{ {H, * } \right\}\) is a subgroup of \(\left\{ {G, * } \right\}\) .
The relation \({R_2}\) is defined for \(a,b \in {\mathbb{Z}^ + }\) by \(a{R_2}b\) if and only if \((4 + \left| {a – b} \right|)\) is the square of a positive integer. Show that \({R_2}\) is not transitive.
Answer/Explanation
Markscheme
(i) Since \({a^2} – {a^2} = 0\) is divisible by n, it follows that \(a{R_1}a\) so \({R_1}\) is reflexive. A1
\(a{R_1}b \Rightarrow {a^2} – {b^2}\) divisible by \(n \Rightarrow {b^2} – {a^2}\) divisible by \(n \Rightarrow b{R_1}a\) so
symmetric. A1
\(a{R_1}b\) and \(b{R_1}c \Rightarrow {a^2} – {b^2} = pn\) and \({b^2} – {c^2} = qn\) A1
\(({a^2} – {b^2}) + ({b^2} – {c^2}) = pn + qn\) M1
so \({a^2} – {c^2} = (p + q)n \Rightarrow a{R_1}c\) A1
Therefore \({R_1}\) is transitive.
It follows that \({R_1}\) is an equivalence relation. AG
(ii) When \(n = 8\) , the equivalence classes are
\(\left\{ {1,3,5,7,9, \ldots } \right\}\) , i.e. the odd integers A2
\(\left\{ {2,6,10,14, \ldots } \right\}\) A2
and \(\left\{ {4,8,12,16, \ldots } \right\}\) A2
Note: If finite sets are shown award A1A1A1.
[11 marks]
Associativity follows since G is associative. A1
Closure: Let \(x,y \in H\) so \(ax = xa\) , \(ay = ya\) for \(a \in G\) M1
Consider \(axy = xay = xya \Rightarrow xy \in H\) M1A1
The identity \(e \in H\) since \(ae = ea\) for \(a \in G\) A2
Inverse: Let \(x \in H\) so \(ax = xa\) for \(a \in G\)
Then
\({x^{ – 1}}a = {x^{ – 1}}ax{x^{ – 1}}\) M1A1
\( = {x^{ – 1}}xa{x^{ – 1}}\) M1
\( = a{x^{ – 1}}\) A1
so \( \Rightarrow {x^{ – 1}} \in H\) A1
The four group axioms are satisfied so \(H\) is a subgroup. R1
[12 marks]
Attempt to find a counter example. (M1)
We note that \(1{R_2}6\) and \(6{R_2}11\) but 1 not \({R_2}11\) . A2
Note: Accept any valid counter example.
The relation is not transitive. AG
[3 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 function \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) is defined by \(\boldsymbol{X} \mapsto \boldsymbol{AX}\) , where \(\boldsymbol{X} = \left[ \begin{array}{l}
x\\
y
\end{array} \right]\) and \(\boldsymbol{A} = \left[ \begin{array}{l}
a\\
c
\end{array} \right.\left. \begin{array}{l}
b\\
d
\end{array} \right]\) where \(a\) , \(b\) , \(c\) , \(d\) are all non-zero.
Consider the group \(\left\{ {S,{ + _m}} \right\}\) where \(S = \left\{ {0,1,2 \ldots m – 1} \right\}\) , \(m \in \mathbb{N}\) , \(m \ge 3\) and \({ + _m}\) denotes addition modulo \(m\) .
Show that \(f\) is a bijection if \(\boldsymbol{A}\) is non-singular.
Suppose now that \(\boldsymbol{A}\) is singular.
(i) Write down the relationship between \(a\) , \(b\) , \(c\) , \(d\) .
(ii) Deduce that the second row of \(\boldsymbol{A}\) is a multiple of the first row of \(\boldsymbol{A}\) .
(iii) Hence show that \(f\) is not a bijection.
Show that \(\left\{ {S,{ + _m}} \right\}\) is cyclic for all m .
Given that \(m\) is prime,
(i) explain why all elements except the identity are generators of \(\left\{ {S,{ + _m}} \right\}\) ;
(ii) find the inverse of \(x\) , where x is any element of \(\left\{ {S,{ + _m}} \right\}\) apart from the identity;
(iii) determine the number of sets of two distinct elements where each element is the inverse of the other.
Suppose now that \(m = ab\) where \(a\) , \(b\) are unequal prime numbers. Show that \(\left\{ {S,{ + _m}} \right\}\) has two proper subgroups and identify them.
Answer/Explanation
Markscheme
recognizing that the function needs to be injective and surjective R1
Note: Award R1 if this is seen anywhere in the solution.
injective:
let \(\boldsymbol{U}, \boldsymbol{V} \in ^\circ \times ^\circ \) be 2-D column vectors such that \(\boldsymbol{AU} = \boldsymbol{AV}\) M1
\({\boldsymbol{A}^{ – 1}}\boldsymbol{AU} = {\boldsymbol{A}^{ – 1}}\boldsymbol{AV}\) M1
\(\boldsymbol{U} = \boldsymbol{V}\) A1
this shows that \(f\) is injective
surjective:
let \(W \in ^\circ \times ^\circ \) M1
then there exists \(\boldsymbol{Z} = {\boldsymbol{A}^{ – 1}}\boldsymbol{W} \in ^\circ \times ^\circ \) such that \(\boldsymbol{AZ} = \boldsymbol{W}\) M1A1
this shows that \(f\) is surjective
therefore \(f\) is a bijection AG
[7 marks]
(i) the relationship is \(ad = bc\) A1
(ii) it follows that \(\frac{c}{a} = \frac{d}{b} = \lambda \) so that \((c,d) = \lambda (a,b)\) A1
(iii) EITHER
let \(\boldsymbol{W} = \left[ \begin{array}{l}
p\\
q
\end{array} \right]\) be a 2-D vector
then \(\boldsymbol{AW} = \left[ \begin{array}{l}
a\\
\lambda a
\end{array} \right.\left. \begin{array}{l}
b\\
\lambda b
\end{array} \right]\left[ \begin{array}{l}
p\\
q
\end{array} \right]\) M1
\( = \left[ \begin{array}{l}
ap + bq\\
\lambda (ap + bq)
\end{array} \right]\) A1
the image always satisfies \(y = \lambda x\) so \(f\) is not surjective and therefore not a bijection R1
OR
consider
\(\left[ {\begin{array}{*{20}{c}}
a&b \\
{\lambda a}&{\lambda b}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
b \\
0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ab} \\
{\lambda ab}
\end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}}
a&b \\
{\lambda a}&{\lambda b}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0 \\
a
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ab} \\
{\lambda ab}
\end{array}} \right]\)
this shows that \(f\) is not injective and therefore not a bijection R1
[5 marks]
the identity element is \(0\) R1
consider, for \(1 \le r \le m\) ,
using \(1\) as a generator M1
\(1\) combined with itself \(r\) times gives \(r\) and as \(r\) increases from \(1\) to m, the group is generated ending with \(0\) when \(r = m\) A1
it is therefore cyclic AG
[3 marks]
(i) by Lagrange the order of each element must be a factor of \(m\) and if \(m\) is prime, its only factors are \(1\) and \(m\) R1
since 0 is the only element of order \(1\), all other elements are of order \(m\) and are therefore generators R1
(ii) since \(x{ + _m}(m – x) = 0\) (M1)
the inverse of x is \((m – x)\) A1
(iii) consider
M1A1
there are \(\frac{1}{2}(m – 1)\) inverse pairs A1 N1
Note: Award M1 for an attempt to list the inverse pairs, A1 for completing it correctly and A1 for the final answer.
[7 marks]
since \(a\), \(b\) are unequal primes the only factors of \(m\) are \(a\) and \(b\)
there are therefore only subgroups of order \(a\) and \(b\) R1
they are
\(\left\{ {0,a,2a, \ldots ,(b – 1)a} \right\}\) A1
\(\left\{ {0,b,2b, \ldots ,(a – 1)b} \right\}\) A1
[3 marks]
Question
Let \(f\) be a homomorphism of a group \(G\) onto a group \(H\).
Show that if \(e\) is the identity in \(G\), then \(f(e)\) is the identity in \(H\).
Show that if \(x\) is an element of \(G\), then \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\).
Show that if \(G\) is Abelian, then \(H\) must also be Abelian.
Show that if \(S\) is a subgroup of \(G\), then \(f(S)\) is a subgroup of \(H\).
Answer/Explanation
Markscheme
\(f(a) = f(ae) = f(a)f(e)\) M1A1
hence \(f(e)\) is the identity in \(H\) AG
\(e’ = f(e)\)
\( = f(x{x^{ – 1}})\) M1
\( = f(x)f({x^{ – 1}})\) A1
hence \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\) AG
let \(a’,{\text{ }}b’ \in H\), we need to show that \(a’b’ = b’a’\) (M1)
since \(f\) is onto \(H\) there exists \(a,{\text{ }}b \in G\) such that \(f(a) = a’\)
and \(f(b) = b’\) (M1)
now \(a’b’ = f(a)f(b) = f(ab)\) A1
since \(f(ab) = f(ba)\) M1
\(f(ba) = f(b)f(a) = b’a’\) A1
hence Abelian AG
METHOD 1
\(e’ = f(e)\) and \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\) from above A1A1
let \(f(a)\) and \(f(b)\) be two elements in \(f(S)\)
then \(f(a)f(b) = f(ab)\) M1
\( \Rightarrow f(a)f(b) \in f(S)\) A1
hence closed under the operation of \(H\)
\(f(S)\) is a subgroup of \(H\) AG
METHOD 2
\(f(S)\) contains the identity, so is non empty A1
Suppose \(f(a),{\text{ }}f(b) \in f(S)\)
Consider \(f(a)f{(b)^{ – 1}}\) M1
\( = f(a)f({b^{ – 1}})\) (from (b)) A1
\( = f(a{b^{ – 1}})\) (homomorphism) A1
\( \in f(S)\) as \(a{b^{ – 1}} \in H\)
So \(f(S)\) is a subgroup of \(H\) (by a subgroup theorem) AG
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]