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 $d = \gcd(p^a, p^b).$
How can I prove the following:
if $a\geq b$ then $f$ is onto.
if $a \leq b$ then $f$ is 1-1.
My intuition is that I will use the fact that I have cyclic groups but how can I relate that to the relation between $a$ and $b,$ could anyone give me a hint please?
To expand on my comment, I'll do $a)$ and leave $b)$ for you to do. The map $$f: \mathbb{Z}/p^a \mathbb{Z} \longrightarrow \mathbb{Z}/p^b \mathbb{Z}$$ is given explicitly by $f(x)=\frac{n}{d}x$. In the first case, this simply sends a number mod $p^a$ to itself mod $p^b$. To see that this is surjective, note that the quotient map $\mathbb{Z} \to \mathbb{Z}/p^b \mathbb{Z}$ factors through $f$ since $a \geq b$.
To do $b)$, you should first ask yourself why $f$ is well-defined. In other words, if $n=m+kp^a$, then why do we have $f(n)=f(m)$? Then try to show that $f(n)=f(m)$ implies $n=m$ in $\mathbb{Z}/p^a \mathbb{Z}$.