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 \(4{ \times _9}x{ \times _9}x = 1\) .

     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]