A non trivial homomorphism from $Q$ to $G$.

group-theory

Suppose the group $(\mathbb{Q},+)$ has a non trivial homomrphism to $G$. Then which of the following can be $G?$

a) $(\mathbb{Z},+)$

b) $(\mathbb{Q}^{\circ},\times)$

c) $(\mathbb{Z}_{m}, +_{(\bmod)})$

d) none of these

For homomorphism, $\frac{\mathbb{Q}}{N}$ should be isomorphic to some subgroup of $G$, where $N$ is normal in $(\mathbb{Q},+)$.

I can take any subgroup of $(\mathbb{Q},+)$ as $N$ as $(\mathbb{Q},+)$ is abelian.

I'm having difficulty in finding the right choice to match with $G$.

Any idea or hint would be helpful.

Best Answer

$\mathbb{Q}$ is a divisible group, meaning every equation of the form $nx=g$ has a solution for $n\in\mathbb{N}$ and $g\in \mathbb{Q}$. You can easily check that a homomorphic image of a divisible group is again divisible. We will use this property to our advantage:

  1. Every subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ except for the trivial group. And since $\mathbb{Z}$ is not divisible (no solution to $2x=1$) then no non-trivial homomorphism $\mathbb{Q}\to\mathbb{Z}$ exists.
  2. $\mathbb{Q}^\circ$ is not divisible. Note that since group operation is multiplication then we are looking for solutions to $x^n=q$. Moreover for any rational $q\neq 1$ there is $n$ such that $x^n=q$ has no solution in $\mathbb{Q}^\circ$. This shows that no subgroup of $\mathbb{Q}^\circ$ is divisible. And so there is no non-trivial homomorphism $\mathbb{Q}\to\mathbb{Q}^\circ$.
  3. $\mathbb{Z}_m$ is not divisible for $m>1$. Indeed, there is no solution to $mx=1$ since $mx=0$ regardless of $x$ (more generally the same argument shows that no finite group is divisible). Since every subgroup of $\mathbb{Z}_m$ is again $\mathbb{Z}_k$ then there is no non-trivial homomorphism $\mathbb{Q}\to\mathbb{Z}_m$.
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