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?

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- May 24th 2011, 01:24 AMdeniselim17intersection of subgroups
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? - May 24th 2011, 01:51 AMSwlabr
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. - May 24th 2011, 04:06 PMdeniselim17
- May 25th 2011, 01:07 AMSwlabr
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.