It is clear that intersection of two normal subgroups of finite index will result a normal subgroup of finite index.
How about intersection of two normal subgroups with prime power index?
Can we get a normal subgroup of some prime power index?
It is clear that intersection of two normal subgroups of finite index will result a normal subgroup of finite index.
How about intersection of two normal subgroups with prime power index?
Can we get a normal subgroup of some prime power index?
Clearly, the indexes must be a power of the same prime!
However, I believe that that is all that is needed.
Let $\displaystyle N_1$ and $\displaystyle N_2$ be the subgroups of finite index, of index $\displaystyle p^{n_1}$ and $\displaystyle p^{n_2}$ respectively. Let $\displaystyle N=N_1\cap N_2$.
Then, if $\displaystyle a^m\in N$ then $\displaystyle a^m\in N_1$ and $\displaystyle a^m\in N_2$. So, clearly $\displaystyle p$ divides $\displaystyle m$. So, let $\displaystyle m=p^nm^{\prime}$ where $\displaystyle p$ does not divide $\displaystyle m^{\prime}$. As $\displaystyle N_1$ and $\displaystyle N_2$ are both of prime power index then $\displaystyle a^{p^n}\in N_1$ and $\displaystyle a^{p^n}\in N_2$. Thus, $\displaystyle a^{p^n} \in N=N_1\cap N_2$ as required.
EDIT: There is a much neater way. One can use the second and third isomorphism theorems!
If $\displaystyle N=N_1\cap N_2$ then $\displaystyle N_2N_1/N_2\cong N_2/N$ by the second isomorphism theorem. This means that $\displaystyle N_2$ is of index $\displaystyle p^m$ for some $\displaystyle m$ (why?).
By the third isomorphism theorem, $\displaystyle \frac{G/N}{N_1/N}\cong G/N_1$. Everything is of finite index, so $\displaystyle [G:N] = [G:N_1]\cdot[N_1:N] = p^{n_1}p^m=p^{n_1+m}$ as required.
For the second isomorphism theorem you need two subgroups, one normal and the other just has to be a subgroup. It doesn't matter if it is normal or not, so long as it is a subgroup. If something holds for a subgroup it will also hold for a normal subgroup, as normal subgroups are just subgroups but with a few more conditions thrown in.
So yes, the second isomorphism theorem can be used in this case.