You only need to show . You can check that , where each g sends each left coset xH to the left coset x'H, and this operation is associative. You might need to pick a representative x from the coset xH and simply say , because if , then .

We next need to show that . We know that and, using the first criterion for an action, we get the required result.No.2). Show that there is a single orbit, , undder this action.

By the Orbit-Stabiliser theorem, I know that .

4). Show that the kernel of the action is a normal subgroup of that is contained in .

I know that .

The elements that satisfy this are going to be all . Therefore, .

(^Is this right?)

This then gives the required result: .

Let M be the set of all left cosets of H in G.

You need to show that if xH and yH are two arbitrary left cosets in M, then you can find an element such that . In this case, choose g as . This means, G acts transitively on M.

Let M be the set of all left cosets of H in G. You need to check the induced homomorphism defined by , where and , where A(M) is the group of permutations of M. If g is in the kernal gxH=xH for all x in G, esp, geH=eH, then g should be in H.I need to show that when .

by substituting in .

I know that is the kernel of the action of on .

Rewriting this as gives that the only element common to every set is when .

The set is indeed a normal subgroup, and must be in since every subgroup of a group shares the same identity element.

The induced homomorphism defined by the previous question can be used here. The kernel of h is contained in H, actually it has the form and it is the largest normal subgroup of G contained in H.

we have left cosets.

THerefore choose to be the subgroup of all left cosets of . This gives a quotient group .

Define a homomorphism by .

I have a proposition in my notes stating that and that is a homomorphism.

From here I get stuck. I've been trying to apply the first isomorphism theorem, but I have no idea how to put it into action.

Anyway, any help would be appreciated.

It follows that G/N is isomorphic to the subgroup of A(M), where A(M) is the group of all permutations of M isomorphic to .

Edit: h is not necessarily a surjective homomorphism. Thus you can't use the first isomorphism theorem here. Rather, you need to show that G/N can be embedded into A(M).