What can I say about a map multiplication?

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:

1. if $a\geq b$ then $f$ is onto.

2. 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 $r= p^t x$ and $0 \leq r < d$? when is this map 1-1 and when it is onto?

Could anyone help me in that please?

Is this map trivially onto?

Answer 1

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 $f[m\mathbb{Z}]=f\left[\langle m \rangle\right]=\left\langle f(m)\right\rangle \leqslant n\mathbb{Z}$, 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 $(k; n)=1$.

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P.S. On the existence and properties of quotient morphisms in the category (context) of groups. I won't fully go into details by proving the statements below (we could do this at a later time, as I confess I currently don't have too much time to spare in order to be active on this forum).

Proposition 1. Let $f \colon G \to G'$ and $g \colon G \to G''$ two group morphisms such that $g$ is surjective and $\mathrm{Ker}g \leqslant \mathrm{Ker}f$. Then there exists a unique group morphism $h \colon G'' \to G'$ such that $f=h \circ g$ (in other words, the diagram below is commutative). We refer to $h$ as the quotient of $f$ with respect to $g$. We have the following descriptions: $$\begin{align} \mathrm{Ker}h&=g\left[\mathrm{Ker}f\right]\\ \mathrm{Im}h&=\mathrm{Im}f. \end{align}$$

Proposition 2 (particular case of the previous). For arbitrary group $G$ and normal subgroup $H \trianglelefteq G$ we adopt the notation $\sigma^G_H \colon G \to G/H$ to refer to the canonical surjection associated to the quotient by $H$. Let $f \colon G \to G'$ be a group morphism together with two normal subgroups $H \trianglelefteq G$ and $H' \trianglelefteq G'$ such that $f[H] \leqslant H'$ (which is equivalent to $H \leqslant f^{-1}\left[H'\right]$). Then there exists a unique morphism $g \colon G/H \to G'/H'$ such that $\sigma^{G'}_{H'} \circ f=g \circ \sigma^G_H$ (in other words, the diagram below is again commutative). We have the descriptions: $$\begin{align} \mathrm{Ker}g&=f^{-1}\left[H'\right]/H\\ \mathrm{Im}g&=\sigma^{G'}_{H'}\left[\mathrm{Im}f\right]=\left(H'\mathrm{Im}f\right)/H'. \end{align}$$

Proposition 2 is obtained from proposition 1 by considering $\sigma^{G'}_{H'} \circ f$ in the role of the original $f$ and $\sigma^G_H$ in the role of the original $g$.