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 $H$, $\quad$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?

 

[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.

 

[blue_raver22]

Hello,

Thank you for your question. Your construction is indeed a specific application of a more general theorem in group theory called the Invariant Subgroup Theorem. This theorem states that for any subgroup H of a group G, there exists a homomorphism \gamma: G \rightarrow G' such that \gamma(h) = e' for all h \in H, where e' is the identity element of G'. In other words, \gamma is an automorphism that fixes the subgroup H.

Your conditions for finding an invariant function w.r.t. the action of H are essentially the same as those required by the Invariant Subgroup Theorem. So, your construction is not only well-known, but it is also a fundamental result in group theory.

I hope this helps answer your question.

Best,