Show that \((G,{\text{ }} + )\) forms a group where \( + \) denotes addition on \(\mathbb{Q}\). Associativity may be assumed.

Assuming that \((H,{\text{ }} + )\) forms a group, show that it is a proper subgroup of \((G,{\text{ }} + )\).

The mapping \(\phi :G \to G\) is given by \(\phi (g) = g + g\), for \(g \in G\).

Prove that \(\phi \) is an isomorphism.

closure: \(\frac{{{n_1}}}{{{6^{{i_1}}}}} + \frac{{{n_2}}}{{{6^{{i_2}}}}} = \frac{{{6^{{i_2}}}{n_1} + {6^{{i_1}}}{n_2}}}{{{6^{{i_1} + {i_2}}}}} \in G\)     A1R1

Note:     Award A1 for RHS of equation. R1 is for the use of two different, but not necessarily most general elements, and the result \( \in G\) or equivalent.

identity: \(0\)     A1

inverse: \(\frac{{ - n}}{{{6^i}}}\)     A1

since associativity is given, \((G,{\text{ }} + )\) forms a group     R1AG

Note:     The R1 is for considering closure, the identity, inverses and associativity.

[5 marks]

it is required to show that \(H\) is a proper subset of \(G\)     (M1)

let \(\frac{n}{{{3^i}}} \in H\)     M1

then \(\frac{n}{{{3^i}}} = \frac{{{2^i}n}}{{{6^i}}} \in G\) hence \(H\) is a subgroup of \(G\)     A1

\(H \ne G\) since \(\frac{1}{6} \in G\) but \(\frac{1}{6} \notin H\)     A1

Note:     The final A1 is only dependent on the first M1.

hence, \(H\) is a proper subgroup of \(G\)     AG

[4 marks]

consider \(\phi ({g_1} + {g_2}) = ({g_1} + {g_2}) + ({g_1} + {g_2})\)     M1

\( = ({g_1} + {g_1}) + ({g_2} + {g_2}) = \phi ({g_1}) + \phi ({g_2})\)     A1

(hence \(\phi \) is a homomorphism)

injectivity: let \(\phi ({g_1}) = \phi ({g_2})\)     M1

working within \(\mathbb{Q}\) we have \(2{g_1} = 2{g_2} \Rightarrow {g_1} = {g_2}\)     A1

surjectivity: considering even and odd numerators     M1

\(\phi \left( {\frac{n}{{{6^i}}}} \right) = \frac{{2n}}{{{6^i}}}\) and \(\phi \left( {\frac{{3(2n + 1)}}{{{6^{i + 1}}}}} \right) = \frac{{2n + 1}}{{{6^i}}}\)     A1A1

hence \(\phi \) is an isomorphism     AG

[7 marks]

Total [16 marks]

This part was generally well done. Where marks were lost, it was usually because a candidate failed to choose two different elements in the proof of closure.

Only a few candidates realised that they did not have to prove that \(H\) is a group - that was stated in the question. Some candidates tried to invoke Lagrange's theorem, even though \(G\) is an infinite group.

Many candidates showed that the mapping is injective. Most attempts at proving surjectivity were unconvincing. Those candidates who attempted to establish the homomorphism property sometimes failed to use two different elements.