Proving Finite Index Subgroups in G Have Normal Subgroups of Lower Index

[CornMuffin]

Homework Statement

Prove: If H is a subgroup with finite index in G
Then there is a normal subgroup K of G such that K is a subgroup of H and K has index less than n! in G.

Homework Equations

Note: |G| represents the index of H in G

|G| is the number of left cosets of H in G, ie # of elements in {gH: g in G}

The Attempt at a Solution

I haven't had much progress in this proof at all.
The only thing that I can think of using is that |G : K| = |G : H||H : K| for a subgroup K of H

But i don't know what to try.

 

[fauboca]

Homework Statement

Prove: If H is a subgroup with finite index in G
Then there is a normal subgroup K of G such that K is a subgroup of H and K has index less than n! in G.

Homework Equations

Note: |G| represents the index of H in G

|G| is the number of left cosets of H in G, ie # of elements in {gH: g in G}

The Attempt at a Solution

I haven't had much progress in this proof at all.
The only thing that I can think of using is that |G : K| = |G : H||H : K| for a subgroup K of H

But i don't know what to try.

Have you tried going from what you have and then using the first iso theorem?

 

[Deveno]

suppose that [G] = n.

consider the mapping φ_g: G/H→G/H given by xH→(gx)H for any given g in G.

prove this mapping is a bijection on G/H, for every g in G (hint: it has an inverse, what is it?).

conclude g→φ_g is a homomorphism of G into Sym(G/H).

what is |Sym(G/H)| (hint: G/H is a set with n elements)?

what can you say about the kernel of the homomorphism g→φ_g?

*****

note that the theorem, strictly speaking, isn't true. for example, let G = S₅, and let H = S₄ = {elements of S₅ that fix 5}. then [G] = 5, but the ONLY normal subgroups of S₅ are A₅ and {e}, and of these two, only {e} is a subgroup of S₄, and {e} has index 5!, in other words, the inequality [G:K] < n! isn't true, but the inequality [G:K] ≤ n! is.