Let \(A=(\mathbb Q,+)\) and let \(M=(\mathbb Q_{>0},\cdot)\). Determine all group homomorphisms \[ A\to M. \]
Solution
Let \[ \varphi:(\QQ,+)\longrightarrow(\QQ_{>0},\cdot) \] be a group homomorphism.
1. For every positive integer \(n\),\[ \varphi(1) = \varphi(1/n)^n. \]
Since \[ 1= \underbrace{\frac1n+\cdots+\frac1n}_{n\text{ terms}}, \] the homomorphism property gives \[ \varphi(1) = \underbrace{\varphi(1/n)\cdots\varphi(1/n)}_{n\text{ factors}} = \varphi(1/n)^n. \]
2. One has\[ \varphi(1)=1. \]
Write the positive rational number \(\varphi(1)\) in its unique prime factorization \[ \varphi(1) = \prod_p p^{e_p}, \] where each \(e_p\in\ZZ\) and all but finitely many \(e_p\) are zero.
By step 1, \(\varphi(1)\) is an \(n\)th power in \(\QQ_{>0}\) for every positive integer \(n\). Therefore every exponent \(e_p\) is divisible by every positive integer \(n\). The only integer with this property is \(0\). Hence every \(e_p=0\), so \[ \varphi(1)=1. \]
3. For every positive integer \(n\),\[ \varphi(1/n)=1. \]
Steps 1 and 2 give \[ \varphi(1/n)^n=1. \] The only positive rational number whose \(n\)th power is \(1\) is \(1\). Hence \[ \varphi(1/n)=1. \]
4. For every rational number \(q\),\[ \varphi(q)=1. \]
Write \[ q=\frac mn \] with \(m\in\ZZ\) and \(n>0\). By step 3, \[ \varphi(1/n)=1. \] If \(m\geq0\), then \[ \varphi(m/n) = \varphi(1/n)^m = 1. \] If \(m<0\), then \[ \varphi(m/n) = \varphi((-m)/n)^{-1} = 1^{-1} = 1. \]
5. The only group homomorphism\[ (\QQ,+)\longrightarrow(\QQ_{>0},\cdot) \]is\[ \boxed{\varphi(q)=1\text{ for every }q\in\QQ}. \]
Step 4 shows that every homomorphism is trivial. Conversely, the constant map \[ q\longmapsto1 \] is plainly a group homomorphism.
6. Q.E.D.
Step 5 gives the complete classification.