homomorphism $\phi$ : $\mathbb Z$ $\rightarrow$ $\mathbb Z$ using $\phi(n)$=$2n$ A question from Visual group theory says :
consider the homomorphism $\phi$ : $\mathbb Z$ $\rightarrow$ $\mathbb Z$ by $\phi(n)$=$2n$,where the group operation on $\mathbb Z$ is '+'.
Would $\phi$ be an homomorphism if it were to use a different coeffiecient than 2?
I think it should remain so for any coeffiecient because it does not violate the properties for homomorphism. Am I correct?
 A: Sure! If you just left $\phi(n)=kn$ then you can show that for all $k \in \mathbb{N}$ that the homomorphism properties hold.
A: To show that $\phi:(G,\ast)\to (G',\star)$ is an homomorphism for all $k \in \mathbb{Z}$, you just need to prove that for all $a,b \in G$:
$$\phi(a \ast b)=\phi(a) \star \phi(b)$$ 
So in the case of the endomorphism $\phi:(\mathbb{Z},+) \to (\mathbb{Z},+)$ with $\phi(n)=kn, k\in\mathbb{Z}$, this means that:
$$k(a+b)=ka+kb$$
A: The group ${\mathbb Z}$ is cyclic, so it suffices to define a homomorphism to another group by giving the image of its generator; we have two generators to choose from, $\pm 1$, so it's perfectly all right to define $\phi$ by giving $\phi(1)$.
${\mathbb Z}$ is free abelian, so any $\phi(1)$ is good.  On the other hand, if you consider ${\mathbb Z}/n{\mathbb Z}$, here the generator must satisfy $n\phi(\overline{1})=0$ for any group homomorphism
$$\phi:{\mathbb Z}/n{\mathbb Z}\to G.$$

In the case where $G={\mathbb Z}/m{\mathbb Z},$ the condition that $\phi(\overline{1})$ must satisfy due to the source group adds to the condition the target group satisfies, i.e.
$$\phi(\overline{1})\in {\mathbb Z}/m{\mathbb Z} \mbox{ and so } m\phi(\overline{1})=0.$$
This is best expressed by the well-known relation 
$$Hom({\mathbb Z}/n{\mathbb Z}, {\mathbb Z}/m{\mathbb Z})={\mathbb Z}/d{\mathbb Z},$$
where $(m,n)=d$.
If $dm_1=m$, then $\phi(1)$ need be a multiple of $m_1$ in ${\mathbb Z}/m{\mathbb Z}$, so both relations satisfy. 

By the above, one always has the relation (in principle, as sets)
$$Hom({\mathbb Z},G)=G$$ given by $\phi \mapsto \phi(1).$  However, the l.h.s. has a group structure only if $G$ is abelian, in general.  If $G$ is abelian, then the above bijection is a group homomorphism.
