Show that the set \(S\) of numbers of the form \({2^m} \times {3^n}\) , where \(m,n \in \mathbb{Z}\) , forms a group \(\left\{ {S, \times } \right\}\) under multiplication.

Show that \(\left\{ {S, \times } \right\}\) is isomorphic to the group of complex numbers \(m + n{\rm{i}}\) under addition, where \(m\), \(n \in \mathbb{Z}\) .

Closure: Consider the numbers \({2^{{m_1}}} \times {3^{{m_1}}}\) and \({2^{{m_2}}} \times {3^{{n_2}}}\) where     M1

\({m_1},{m_2},{n_1},{n_2}, \in \mathbb{Z}\) . Then,

Product \( = {2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}}\) which \( \in S\)     A1

Identity: \({2^0} \times {3^0} = 1 \in S\)     A1

Since \(({2^m} \times {3^n}) \times ({2^{ - m}} \times {3^{ - n}}) = 1\) and \({2^{ - m}} \times {3^{ - n}} \in S\)     R1

then \({2^{ - m}} \times {3^{ - n}}\) is the inverse.     A1

Associativity: This follows from the associativity of multiplication.     R1

[6 marks]

Consider the bijection

\(f({2^m} \times {3^n}) = m + n{\rm{i}}\)    (M1)

Then

\(f({2^{{m_1}}} \times {3^{{n_1}}}) \times ({2^{{m_2}}} \times {3^{{n_2}}}) = f({2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}})\)     M1A1

\( = {m_1} + {m_2} + ({n_1} + {n_2}){\rm{i}}\)     A1

\( = ({m_1} + {n_1}{\rm{i}}) + ({m_2} + {n_2}{\rm{i}})\)     (A1)

\( = f({2^{{m_1}}} \times {3^{{n_1}}}) + f({2^{{m_2}}} \times {3^{{n_2}}})\)     A1

[6 marks]