# why there does not exist any homomorphism f : Q → Qpos.

If G is a group and b ∈ G, then a square root of b is an element a ∈ G such that a^2= b. For example 3 is a square root of 6 in 〈Z,+>

Let 〈Q,+> be the group of rational numbers with the operation of addition and let 〈Qpos, . >be the group of positive rational numbers with the operation of multiplication. Use the fact that you proved in part a to explain why there does not exist any homomorphism f : Q → Qpos.

SO FAR , I know let f be homomorphism from 〈Q,+> to　〈Qpos, . >defined by f(x+y) = x times y, let x = 0, and y = 1 we will get f(1)= 0 but 0 is not in Qpos. do you think my method is right? because I didn't use square root in this part of question...

the part a is about {a. Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Prove that if every element of G has a square root then every element of H also has a square root.}

• Qpos is the second group? and do you mean no non-trivial hom? – GiantTortoise1729 Dec 4 '15 at 3:55
• What have you tried? Where did you get stuck? (Also, what did you prove in part a?) EDIT: All that you've shown is that one specific map is not a homomorphism; you need to show that there is no homomorphism, at all. – Noah Schweber Dec 4 '15 at 3:55
• @NoahSchweber that part a is (a. Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Prove that if every element of G has a square root then every element of H also has a square root.) I have already solve it. – Helen Dec 4 '15 at 3:58
• @GiantTortoise1729 I edit my question, Thank you! – Helen Dec 4 '15 at 4:00

There is such a homomorphism, namely the trivial homomorphism where everything is mapped to $1$. However this is the only possibility.
Proof. In $\def\Q{{\Bbb Q}}\Q$ you can take the square root of any element. Repeating, you can take the square root of the square root of the square root of... the square root of any element, with as many square roots as you like. If there is a homomorphism $\phi$ from $\Q$ to $\Q^+$, then the same thing holds in $R={\rm range}\,\phi$. But in $R$, square root has its normal meaning, and given any element $a$ of $\Q^+$, the only way you can take the square root of the square root... indefinitely many times and always get rational results is if $a=1$. Therefore ${\rm range}\,\phi=\{1\}$, and $\phi$ is the trivial homomorphism.