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Prove that if H has finite index n then there is a normal subgroup N of G with $N \subset H $ and [G:N]$ \le $n!.

I tried to solve the problem but could not done exactly. Since [G:H]=n , Let A={$a_i H$:i=1,2,..,n}. Consider the left multiplication group action on A .Then kernal of the action is intersection N= $\cap a_iHa_i^{-1}$ i=1,2..,n. Then N is the largest normal subgroup of G contained in H. Now it is clear that $[G:a_iHa_i^{-1}]$ =n for all i.{by orbit stabliser theorem}.Now how to find [G:N]? yes I got it. By Caley's theorem G/N is isomorphic to a subgroup of $ S_{A}$ .Therefone [G:N] divides n!.

Digjoy Paul
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Define a homomorphism $F$ from $G$ to the group $\{f: G/H \to G/H| f \mbox{ is bijective}\}$ and look at its kernel ($N$ is to be related with it).

  • Here H need not be normal. So is G/H well defined? – Digjoy Paul Sep 18 '14 at 05:04
  • Instead of this you can have homomorphism between G and $ S_A $ ,the associative homomorphism corresponding to the left multiplicative group action on the set A,set of distinct coset of H. – Digjoy Paul Sep 18 '14 at 05:17
  • Never did I say that G/H was a group and we actually don't need that it is one. Here G/H is just the set of distinct (right) cosets of H in G. –  Sep 18 '14 at 06:03
  • @J.H.S. The problem is that many group theorists only use the notation $G/H$ when $H$ is a normal subgroup of $G$, so to use it otherwise is likely to cause confusion. – Derek Holt Sep 18 '14 at 07:18
  • Prof. Derek: Was the notation G/H the standard way to refer to set of (right) cosets of H in G at some point of time? –  Sep 19 '14 at 08:09