$\mathbb G = \mathbb Z_{p^3} \oplus\mathbb Z_{p^3}\oplus\mathbb Z_{p}$ - it's group, p - prime.
Find the number of decompositions of $\mathbb G$ into a direct sum of cyclic groups.
I found the number of elements in a group: $p^3$ ---> $p^7 - p^6$ elements; $p^2$ ---> $p^6 - p^3$ elements; $p$ ---> $p^3 - 1$ elements;
Then i did this: $\mathbb G = \mathbb A \oplus\mathbb B\oplus\mathbb C$ . Where exponent of A = $p^3$, exp of B = $p^3$ and exp of C = $p$.
For A we have $ (p^7 - p^6)/(p^3 - p^2) = p^4 $ options. For C we have $(p^3 - 1 - p^2 + 1)/(p-1) = p^2$ options. For $\mathbb B$, we must find the size of intersections of groups $\mathbb Z_{p^3}$ and $\mathbb Z_{p^3}$ in groups $\mathbb Z_{p^2}$ and $\mathbb Z_p$. And then do the same as in $\mathbb A$.
Result will be: $p^4 * p^2 * x$, where x = the number of options for B..
Question: how i can find the size of intersections of groups? If I have any mistakes, please tell me.
P.S. Sorry for my bad English.