Define a group homomorphism $\phi: \frac{\mathbb{Z}}{p^{r_1}\mathbb{Z}} \oplus ... \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}} \to \frac{\mathbb{Z}}{p^{r_1}\mathbb{Z}} \oplus ... \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}}$ by seinding each element to its multiple by $p$( $p$ is a prime number and $r_1 \geq ... \geq r_m$ ).
What is the image $p(\frac{\mathbb{Z}}{p^{r_1}\mathbb{Z}} \oplus ... \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}})$ of such homomorphism? How can we characterize this group( the image )?
Of course, each element of it will have the form $(n_1[p]_{p^{r_1}}, ..., n_m[p]_{p^{r_m}} )$ for some $n_i \in \{0, ..., p^{r_i} - 1 \}$.
But there should be more to it( I know since it's a part of an exercise ).
It appears that I have understood the issue. First of all, $p(\frac{\mathbb{Z}}{p^{r_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}}) = \langle ([p]_{p^{r_1}}, ..., [p]_{p^{r_m}}) \rangle$.
I claim that $\langle ([p]_{p^{r_1}}, ..., [p]_{p^{r_m}}) \rangle \cong \frac{\mathbb{Z}}{p^{r_1 - 1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m - 1} \mathbb{Z}}$.
Define $\phi: \langle ([p]_{p^{r_1}}, ..., [p]_{p^{r_m}}) \rangle \to \frac{\mathbb{Z}}{p^{r_1 - 1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m - 1} \mathbb{Z}}$ by
$\phi(n_1[p]_{p^{r_1}}, ..., n_m[p]_{p^{r_m}}) = ([n_1]_{p^{r_1 - 1}}, ..., [n_m]_{p^{r_m - 1}})$
It is well-defined as a function:
$[n_ip]_{p^{r_i}} = [n'_ip]_{p^{r_i}} \Leftrightarrow n_ip \equiv n'_ip \mod p^{r_i} \Leftrightarrow p(n_i - n'_i) = kp^{r_i} \Leftrightarrow n_i- n'_i = kp^{r_i - 1} \Leftrightarrow n_i \equiv n'_i \mod p^{r_i - 1} \Leftrightarrow [n_i]_{p^{r_i - 1}} = [n'_i]_{p^{r_i - 1}}$
Hence,
$(n_1[p]_{p^{r_1}}, ..., n_m[p]_{p^{r_m}}) = (n'_1[p]_{p^{r_1}}, ..., n'_m[p]_{p^{r_m}}) \Rightarrow \phi(n_1[p]_{p^{r_1}}, ..., n_m[p]_{p^{r_m}}) = \phi(n'_1[p]_{p^{r_1}}, ..., n'_m[p]_{p^{r_m}})$
So, the function is well-defined.
It is a homomorphism:
$\phi((n_1[p]_{p^{r_1}}, ..., n_m[p]_{p^{r_m}}) + (n'_1[p]_{p^{r_1}}, ..., n'_m[p]_{p^{r_m}})) = \phi((n_1+n'_1)[p]_{p^{r_1}}, ..., (n_m+n'_m)[p]_{p^{r_m}}) = ([n_1+n'_1]_{p^{r_1-1}}, ..., [n_m - n'_m]_{p^{r_m-1}}) = ([n_1]_{p^{r_1-1}}, ..., [n_m]_{p^{r_m}-1}) + ([n'_1]_{p^{r_1-1}}, ..., [n'_m]_{p^{r_m}-1}) = \phi(n_1[p]_{p^{r_1}}, ..., n_m[p]_{p^{r_m}}) + \phi(n'_1[p]_{p^{r_1}}, ..., n'_m[p]_{p^{r_m}})$
It's also injective: $[n_i]_{p^{r_i-1}} = [n'_i]_{p^{r_i-1}} \Leftrightarrow n_i \equiv n'_i \mod p^{r_i-1} \Leftrightarrow p(n_i - n'_i) = kp^{r_i} \Leftrightarrow pn_i - pn'_i = kp^{r_i} \Leftrightarrow pn_i \equiv pn'_i \mod p^{r_i} \Leftrightarrow [pn_i]_{p^{r_i}} = [pn'_i]_{p^{r_i}}$
And surjective:
$\forall n_1, ..., n_m \ \ ([n_1]_{p^{r_1-1}}, ..., [n_m]_{p^{r_m}-1}) = \phi(p[n_1]_{p^{r_1-1}}, ..., p[n_m]_{p^{r_m}-1})$.
So, $\phi$ is an isomorphism, and $p(\frac{\mathbb{Z}}{p^{r_1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m} \mathbb{Z}}) \cong \frac{\mathbb{Z}}{p^{r_1 - 1} \mathbb{Z}} \oplus ... \oplus \frac{\mathbb{Z}}{p^{r_m - 1} \mathbb{Z}}$.