Determine if there exists a homomorphism $f:(\Bbb Z, +) \to (\Bbb Q, +)$ for which $f(1) = 1$. Similarly determine if there exists a homomorphism $g:(\Bbb Q, +) \to (\Bbb Z, +)$ for which $g(1) = 1$.
Wouldn’t both these be satisfied by the identity map $x \mapsto x$? If $-1, 2 \in \Bbb Z$, then $f(-1+2) = f(-1)+f(2) = -1+2=1=f(1)$?
Also take $\frac34, \frac14 \in \Bbb Q$. Now $g(\frac34+\frac14)=g(\frac34)+g(\frac14)=1=g(1)$?
Am I missing somehthing here? Isn’t the identity a homomorphism?
You can define $f\colon\Bbb Z\longrightarrow\Bbb Q$ by $f(x)=x$, and, yes, it maps $1$ into $1$ and it is a homomorphism from $(\Bbb Z,+)$ into $(\Bbb Q,+)$.
But it makes no sense to talk about the map $g\colon\Bbb Q\longrightarrow\Bbb Z$ defined by $g(x)=x$. What would be $g\left(\frac12\right)$ then?
Actually, there is no group homomorphism from $(\Bbb Q,+)$ into $(\Bbb Z,+)$ other than the null function. In fact, if $g$ is such a homomorphism and if $g(1)\ne0$, then take $n\in\Bbb N$ such that $n\nmid g(1)$. Then$$g(1)=g\left(n\times\frac1n\right)=ng\left(\frac1n\right),$$which is impossible, since $n\nmid g(1)$. So, $g(1)=0$. Now, if $m\in\Bbb Z$, then $g(m)=mg(1)=0$. And finally, if $n\in\Bbb N$, $ng\left(\frac mn\right)=g(m)=0$, and therefore $g\left(\frac mn\right)=0$.