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 and be the subgroups of finite index, of index and respectively. Let .
Then, if then and . So, clearly divides . So, let where does not divide . As and are both of prime power index then and . Thus, as required.
EDIT: There is a much neater way. One can use the second and third isomorphism theorems!
If then by the second isomorphism theorem. This means that is of index for some (why?).
By the third isomorphism theorem, . Everything is of finite index, so 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.