Exercise: Let $G$ be a finite group, find the number of homomorphisms: i)$f:\mathbb{Z}\to G$. ii) $f:G\to\mathbb{Z}$
Solution: i) As $\mathbb{Z}=\langle 1\rangle$ is an infinite cyclic group, then for each $a\in G$ there is only one homomorphism $f:\mathbb{Z}\to G$ such that $f(1)=a$, therefore there are as many homeomorphisms as the number of elements in $G$ ii) As $G$ has finite order then any element $a\in G$ must have a finite order and since $f:G\to\mathbb{Z}$ is a homomorphism, $f(a)$ has finite order, therefore there is an unique trivial homomorphism $f:G\to\mathbb{Z}$, so that $f(a)=0$.
I think the author´s solution is wrong because the homomorphism should be f(0)=a, once if it was f(1)=a, the order of $1$ would be infinite. According to ii), the author must be referring to the group $\mathbb{Z},+$ as in i), the author may be referring to $(\mathbb{Z}-{0},.)$ however as it is stated $\mathbb{Z}$ we must assume the operation is the sum.
Question:
Is my line of thinking right? Is the solution wrong?
Thanks in advance!