Question about invariant w.r.t. a group action

[mnb96]

Hello,

I have a group (G,\cdot) that has a subgroup H \leq G, and I consider the action of H on G defined as follows:
\varphi(h,g)=h\cdot g
In other words, the action is simply given by the group operation.

Now I am interested in finding a (non-trivial) invariant function w.r.t. the action of H, which means finding a function \chi:G\rightarrow G' such that \chi(h\cdot g)=\chi(g) for all h\in H and g\in G.

I realized that I can easily impose sufficient conditions on \chi to ensure that it is an invariant function w.r.t. H.
Such conditions are:

1. \chi has the form \chi(g) = \gamma(g)^{-1}\cdot g
   
2. \gamma is an automorphism of the group G
3. \gamma fixes the subgroup H, \quadi.e. \gamma(h)=h for all h\in H

The proof is very easy (just apply \chi as defined in 1. to (h\cdot g) and use 2. and 3.)

My question is: Are the above conditions already well-known, perhaps in a more general form?
Is my construction just a specific application of some more general theorem in group theory?

 

[lavinia]

Your three conditions seem to not work but maybe I don't understand what you are saying.

In any case,,any group homomorphism whose kernel contains the normalizer of H would seem to work.

 

[mnb96]

Hi Lavinia,

thanks for replying. I was wondering why you said that the three conditions I listed do not work. The proof went like this:

\chi(hg)= \gamma(hg)^{-1}hg \quad\quad (definition of \chi)
= \gamma(g)^{-1} \gamma(h)^{-1} hg \quad\quad (\gamma is an automorphism)
= \gamma(g^{-1}) \gamma(h^{-1}) hg \quad\quad(\gamma is an automorphism)
= \gamma(g^{-1}) h^{-1}hg \quad\quad(\gamma fixes H)
= \gamma(g^{-1}) g
= \chi(g)

Did I miss something? I think I have used all the three assumptions mentioned in my previous post.

I feel that what I have done here is essentially the consequence of some well-known construction/theorem in group theory, but I don't know which one. I didn't get completely your argument about the normalizer.

 

[lavinia]

Your calculation is correct. I was confused.

A homomorphism whose kernel contains H will satisfy your required condition since all of the elements of H are sent to the identity. The normalizer is automatically in the kernel.