Thread: proof that subgroup has finite index

1. proof that subgroup has finite index

Q) If H is of finite index in G prove that there exists a subgroup N of G contained in H, and of finite index in G such that $aNa^{-1} = N$ .

In a previous problem, I proved that if H is a subgroup of G, then $N = \bigcap xHx^{-1}$ is a subgroup of G such that $aNa^{-1} = N$. So I'm assuming that this is the same subgroup that is being talked about in the above question. How do I prove that it has finite index?

2. Originally Posted by sashikanth
Q) If H is of finite index in G prove that there exists a subgroup N of G contained in H, and of finite index in G such that $aNa^{-1} = N$ .

In a previous problem, I proved that if H is a subgroup of G, then $N = \bigcap xHx^{-1}$ is a subgroup of G such that $aNa^{-1} = N$. So I'm assuming that this is the same subgroup that is being talked about in the above question. How do I prove that it has finite index?
Because $H$ has finite index we have that there are only finitely many of these $xHx^{-1}$, and each has finite index.

It then suffices to prove that the intersection of finitely many cosets (or whatever these thing are called) of finite index is a coset of finite index. Use induction to prove this. (Although be warned, I wouldn't call this proof "easy"...)

3. Originally Posted by Swlabr
Because $H$ has finite index we have that there are only finitely many of these $xHx^{-1}$, and each has finite index.
What is the proof that each of the $xHx^{-1}$ has finite index? I'm sorry if its a trivial question, but I can't seem to hit the nail on the head!

4. Originally Posted by sashikanth
What is the proof that each of the $xHx^{-1}$ has finite index? I'm sorry if its a trivial question, but I can't seem to hit the nail on the head!
Why isn't the mapping $\lambda:H\mapsto xHx^{-1}$ a surjection?

5. Originally Posted by Drexel28
Why isn't the mapping $\lambda:H\mapsto xHx^{-1}$ a surjection?
It is surjective. Let me try and answer my question now. The mapping $\lambda:H\mapsto xHx^{-1}$ is onto, I think it is actually a bijection. So the number of elements in $H$ equal the number of elements in $xHx^{-1}$. Since the size of the coset equals the order of the subgroup, the size of the cosets of $xHx^{-1}$ equal the size of the cosets of $H$. Since cosets span the group, their index has to be equal.

What if the order of the subgroup H is not finite?

6. Originally Posted by sashikanth
It is surjective. Let me try and answer my question now. The mapping $\lambda:H\mapsto xHx^{-1}$ is onto, I think it is actually a bijection. So the number of elements in $H$ equal the number of elements in $xHx^{-1}$. Since the size of the coset equals the order of the subgroup, the size of the cosets of $xHx^{-1}$ equal the size of the cosets of $H$. Since cosets span the group, their index has to be equal.

What if the order of the subgroup H is not finite?
If $N=\bigcap_{x \in G}xHx^{-1}$, then it is the kernel of $h:G \rightarrow A(M)$, where A(M) is the group of permutations of M and M is the set of all left cosets H in G. Here, h is the induced homomorphism of the group G acting on left cosets of H in G (see here).
Since G/N is isomorphic to the subgroup of A(M), the index of N in G should be finite.

7. Thank you for your reply. I have just started studying groups, I'm on section 2.5 of Topics in Algebra. I dont yet understand isomorphisms. I'll check this solution after I've covered the same.