Show that {T, ×₁₀} is a group. (You may assume associativity.)

By making reference to the Cayley table, explain why T is Abelian.

Find the order of each element of {T, ×₁₀}.

Hence show that {T, ×₁₀} is cyclic and write down all its generators.

The binary operation multiplication modulo 10, denoted by ×₁₀ , is defined on the set V = {1, 3 ,5 ,7 ,9}.

Show that {V, ×₁₀} is not a group.

closure: there are no new elements in the table      A1

identity: 6 is the identity element      A1

inverse: every element has an inverse because there is a 6 in every row and column (2⁻¹ = 8, 4⁻¹ = 4, 6⁻¹ = 6, 8⁻¹ = 2)      A1

we are given that (modulo) multiplication is associative      R1

so {T, ×₁₀} is a group      AG

[4 marks]

the Cayley table is symmetric (about the main diagonal)      R1

so T is Abelian      AG

[1 mark]

considering powers of elements      (M1)

     A2

Note: Award A2 for all correct and A1 for one error.

[3 marks]

EITHER

{T, ×₁₀} is cyclic because there is an element of order 4      R1

Note: Accept “there are elements of order 4”.

OR

{T, ×₁₀} is cyclic because there is generator      R1

Note: Accept “because there are generators”.

THEN

2 and 8 are generators      A1A1

[3 marks]

EITHER

considering singular elements      (M1)

5 has no inverse (5 ×₁₀ a = 1, a∈V has no solution)      R1

OR

considering Cayley table for {V, ×₁₀}

     M1

the Cayley table is not a Latin square (or equivalent)      R1

OR

considering cancellation law

eg, 5 ×₁₀9 = 5 ×₁₀ 1 = 5      M1

if {V, ×₁₀} is a group the cancellation law gives 9 = 1      R1

OR

considering order of subgroups

eg, {1, 9} is a subgroup      M1

it is not possible to have a subgroup of order 2 for a group of order 5 (Lagrange’s theorem)      R1

THEN

so {V, ×₁₀} is not a group     AG

[2 marks]