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

1. Sep 23, 2014

### 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 $H$, $\quad$i.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?

2. Oct 6, 2014

### 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.

3. Oct 6, 2014

### 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.

4. Oct 6, 2014

### 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.