I have asked this question here What can I say about a map multiplication?, here is the question and its answer:
The question:
If I know the following:
If I have the following homomorphism $f: \mathbb Z_{p^a} \rightarrow \Bbb Z_{p^b}$ defined by multiplication by $n/d$ where $n = p^b$ and $d = \gcd(p^a, p^b).$
And I know the following:
if $a\geq b$ then $f$ is onto.
if $a \leq b$ then $f$ is 1-1.
How can I use this piece of information to conclude something about this function: multiplication by $rn/d$ for $r= p^s x$ and $0 \leq r < d$? when is this map 1-1 and when it is onto?
Could anyone help me in that please?
Its Answer:
Let us handle the natural generalisation of the problem you present. Consider $m, n \in \mathbb{N}^{\times}$ together with the notation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{Z}$. Let $d\colon=(m; n)$ denote the greatest common divisor of the two numbers and also consider $m'\colon=\frac{m}{d}, n'=\frac{n}{d}$ (the fractions exist since $d \neq 0$).
Let us fix $k \in \mathbb{N}^{\times} \cap n'\mathbb{Z}$ and let $f \in \mathrm{End}_{\mathbf{Gr}}(\mathbb{Z})$ be the additive group endomorphism given by $f(r)=kr$ for arbitrary $r \in \mathbb{Z}$. It is obvious that $f[\mathbb{Z}]=f\left[\langle m \rangle\right]=\left\langle f(m)\right\rangle \leqslant n\mathbb{Z}$, since $f(m)=mk \in mn'\mathbb{Z}=m'dn'\mathbb{Z}=m'n\mathbb{Z} \leqslant n\mathbb{Z}$, which means that $f$ induces a quotient morphism $g \in \mathrm{Hom}_{\mathbf{Gr}}\left(\mathbb{Z}_m, \mathbb{Z}_n\right)$, described by $g\left(\overline{r}\right)=\widehat{f(r)}=\widehat{kr}$, where $\overline{\bullet}$ denotes classes modulo $m$ respectively $\widehat{\bullet}$ refers to classes modulo $n$.
By virtue of the general properties of quotient morphisms, we have $\mathrm{Ker}g=f^{-1}\left[n\mathbb{Z}\right]/m\mathbb{Z}$. It is easy to see that $f^{-1}\left[n\mathbb{Z}\right]=\frac{n}{(k; n)}\mathbb{Z}$, whence $\mathrm{Ker}g=\frac{n}{(k; n)}\mathbb{Z}/m\mathbb{Z} \approx \mathbb{Z}_{\frac{m(k;n)}{n}} \ (\mathbf{Gr})$. In particular, $g$ is injective if and only if $\frac{m(k; n)}{n}=1$, relation equivalently expressed as $m(k; n)=n \Leftrightarrow m \mid n \wedge (k; n)=\frac{n}{m}$.
As to the image, we have that $\mathrm{Im}g=\left(\mathrm{Im}f+n\mathbb{Z}\right)/n\mathbb{Z}=\left(k\mathbb{Z}+n\mathbb{Z}\right)/n\mathbb{Z}=(k; n)\mathbb{Z}/n\mathbb{Z}=\frac{n}{(k; n)}\mathbb{Z} \ (\mathbf{Gr})$. Therefore, $g$ is surjective if and only if $(k; n)=1$, which entails $n'=1$ and thus $n \mid m$. The surjectivity of $g$ is thus attained if and only if $n \mid m$ and $(k; n)=1$.
Here is my question now:
I have a case that contradicts the required conditions for $g$ to be being surjective If we took $m=2^4, n = 2^3 $ and $r = 2$ that case will not be surjective. Am I correct in that example or no? could anyone help me in that?
From what I can tell, the assertion is that the group homomorphism $f : \mathbb Z \to \mathbb Z$ defined by $f(r) = kr$ induces a surjective group homomorphism $g(r + m \mathbb Z) = kr + n \mathbb Z$ if and only if $n \,|\, m$ and $\gcd(k, n) = 1.$ For the case that $m = 2^4,$ $n = 2^3,$ and $r = 2,$ we have that $$k = \frac{rn}{\gcd(m, n)} = \frac{2^4}{2^3} = 2$$ so that $\gcd(k, n) = 2.$ Ultimately, there is no contradiction because $\gcd(k, n) \neq 1.$