(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.

(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]

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.