Mapping Generators to Generators

Question

I am trying to determine for which integers $z$ the map $f_z : \mathbb{Z}/48\mathbb{Z} \rightarrow \Bbb{Z}_{36}$ with $f_z(\overline{1}) = x^a$, where $x$ is the generator of $\Bbb{Z}_{36}$, extends to homomorphism. I want to say that $f$ is a homomorphism only if it maps $\overline{1}$ to some generator in $\Bbb{Z}_{36}$. But that would require that something like the following is true:

Let $G$ and $H$ be cyclic groups of order $n$ and $m$, reps., with generators $g$ and $h$, reps. Then $f : G \rightarrow H$ is a homomorphism iff one of the generator in $G$ gets mapped to another in $H$.

Now, I imagine the claim as I presented it isn't true; it certainly isn't articulated very well. So, my question is, is there some theorem like the above that would prove useful for my problem; i.e., what other stipulations would we have to add?

EDIT:

Pmar seems to be claiming that $f_z$ need not map $\overline{1}$ to some generator, yet Bob Wilson's answer here, which has $3$ up-votes mind you, seems to suggest otherwise. Would someone mind addressing this ostensible discrepancy?

Answer 1

I'll give a general answer: if $p, q$ are integers, $d=\gcd(p,q)$, set $p=p'd$, $q=q'd$. We have $$\DeclareMathOperator{\Hom}{Hom}\def\Z{\mathbf Z}\Hom(\Z/p\Z,\Z/q\Z)\simeq \Z/d\Z. $$

Indeed, for any abelian group $G$, $\Hom(\Z/p\Z,G)\simeq 0:_G p\;$ (the annihilator of $p$ in $G$). If $G=\Z/q\Z$, this yields $$\Hom(\Z/p\Z,\Z/q\Z)\simeq 0 :_{\Z/q\Z}p=q'\Z/q\Z=q'\Z/q'd\Z\simeq\Z/d\Z. $$

Thus, for the problem at hand, a generator of $\Z/48\Z$ must be mapped onto a multiple of $3\times{}$a generator of $\Z/36\Z$.

Answer 2

If by 'another in H' you mean 'a generator of all of H', then you are correct, the proposition is false. What is true is that the image of any generator of G generates a subgroup of H, but there is no guarantee that the image would generate all of H.

Take your given example: The generator 1 has order 48 in G, so the homomorphic property implies $f(1^{48})$ = $f(1)^{48}$ = 0 in H; but we also have $f(1)^{36}$ = 0 in H; hence $f(1)^{12}$ = 0 in H, so $f(1)$ cannot generate all of H.