# intersection of subgroups

• May 24th 2011, 01:24 AM
deniselim17
intersection 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 AM
Swlabr
Quote:

Originally Posted by deniselim17
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 $N_1$ and $N_2$ be the subgroups of finite index, of index $p^{n_1}$ and $p^{n_2}$ respectively. Let $N=N_1\cap N_2$.

Then, if $a^m\in N$ then $a^m\in N_1$ and $a^m\in N_2$. So, clearly $p$ divides $m$. So, let $m=p^nm^{\prime}$ where $p$ does not divide $m^{\prime}$. As $N_1$ and $N_2$ are both of prime power index then $a^{p^n}\in N_1$ and $a^{p^n}\in N_2$. Thus, $a^{p^n} \in N=N_1\cap N_2$ as required.

EDIT: There is a much neater way. One can use the second and third isomorphism theorems!

If $N=N_1\cap N_2$ then $N_2N_1/N_2\cong N_2/N$ by the second isomorphism theorem. This means that $N_2$ is of index $p^m$ for some $m$ (why?).

By the third isomorphism theorem, $\frac{G/N}{N_1/N}\cong G/N_1$. Everything is of finite index, so $[G:N] = [G:N_1]\cdot[N_1:N] = p^{n_1}p^m=p^{n_1+m}$ as required.
• May 24th 2011, 04:06 PM
deniselim17
Quote:

Originally Posted by Swlabr
If $N=N_1\cap N_2$ then $N_2N_1/N_2\cong N_2/N$ by the second isomorphism theorem. This means that $N_2$ is of index $p^m$ for some $m$ (why?).

By the third isomorphism theorem, $\frac{G/N}{N_1/N}\cong G/N_1$. Everything is of finite index, so $[G:N] = [G:N_1]\cdot[N_1:N] = p^{n_1}p^m=p^{n_1+m}$ as required.

2nd Isomorphism Theorem can be used in this case?
I thought this theorem can only be used when one of the subgroup is normal, not both normal.
• May 25th 2011, 01:07 AM
Swlabr
Quote:

Originally Posted by deniselim17
2nd Isomorphism Theorem can be used in this case?
I thought this theorem can only be used when one of the subgroup is normal, not both normal.

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.