Homomorphism from $(\Bbb Q,+)$ to a finite group 
Prove that if $f$ is a homomorphism from $(\Bbb Q,+)$ to a finite group $G$ then $f(q)=e_G$ for all $q$ in $\Bbb Q$. 

I attempted the following:
Firsty I reason that $f(1)$ generates the entire image, because $f(p/q)=p/qf(1)$. But since $f[\Bbb Q]$ is an infinite subgroup of $G$ and $G$ itself is finite we are forced to let $f(1)=e_G$.
Is this ok?

The only cyclic groups are (to isomorphic groups) $\Bbb{Z}$
infinite cyclic group  and $\Bbb{Z}/n\Bbb{Z},\; n\in\Bbb{N}$
finite cyclic group of order $n$, so $\Bbb{Q}$ not cyclic.
if $f$ is homomorphism of groups from $\Bbb{Q}$ to $G$ then in
particular
 $f(p+q)=f(p)^q=f(q)^p=f(1)^{p+q} \forall (p,q)\in \Bbb{Z}\times
\Bbb{Z}$ and $f(p\frac{p'}{q'})=f(\frac{p'}{q'})^p$  (here we
denote the low of $G$ as multiplicative), so if $n=|G|$ we have
for all $m$ in $\Bbb{Z}$,  $\; f(m)=f(\frac{nm}{n})=(f(\frac
mn))^n=e_G$ (by Lagrange theorem);  in particular   $f(1)=e_G$.
for $m$ not zero $f(\frac 1m)=f(\frac
{n}{nm})=(f(\frac{1}{nm}))^n=e_G$ (also by lagrange theorem).
finally we have $f(\frac pq)=e_G \; \forall (p,q)\in \Bbb{Z}\times
\Bbb{Z}^*$.
