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 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]
Examiners report
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]
Examiners report
(a)(i) This was routine start to the question, but some candidates thought that commutativity was necessary as a group property.
(ii) Showing why 1 and 5 were generators would have been appropriate since this is needed for the cyclic property of the group.
(ii) This did not prove difficult for most candidates.
There were some long, confused arguments that did not lead anywhere. Candidates often do not appreciate the significance of “if” and “only”.
There were some long, confused arguments that did not lead anywhere. Candidates often do not appreciate the significance of “if” and “only”.
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]
Examiners report
Most candidates drew a table for this part and generally achieved success in both (i) and (ii).
In (b) most did use Cayley tables and managed to match element order but could not clearly state the two possible bijections. Sometimes showing that the two groups were isomorphic was missed.
Part B was not well done and the properties of a three element group were often quoted without any proof.
Part B was not well done and the properties of a three element group were often quoted without any proof.