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
(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
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]