category-theory, right group action

Question

Let $G$ be a group. We observe the category $(Set)_G$ of right group actions.

a) Let $X\times G\to X$ and $Y\times G\to Y$ be two transitive right group actions with $x\in X$ and $y\in Y$. Find a criterion, when it exists a morphism $f:X\to Y$ of $(Set)_G$ with $f(x)=y$. Constitute your answer.

b) Conclude: Every transitive group action $X\times G\to X$ is isomorphic in $(Set)_G$ to the action $_H/G\times G\to _H/G\,\,, (Hg,g')\mapsto Hgg'$ for a suitable subgroup $H$ of $G$.

Hello,

I have a question to this task. I want to find the criterion, which is asked for. The morphisms of $(Set)_G$, are functions which hold $f(xg)=f(x)g$ for every $x,g$, and the objects are the right group actions.

I have to misunderstand something. $f:X\to Y$ is supposed to be a morphism of $(Set)_G$. That means $X,Y$ are objects of the category $(Set)_G$. The objects are right group actions, but $X,Y$ are sets, and not functions.

The morphism $f$ has to satisfy some conditions. I would assume to have such an $f$ and then try to find the connection to the given transitive right group actions to come up with a criterion, when this $f$ exists. But first I have to understand how the morphism $f$ "works".

Could you clarify, what I seem to misunderstand here? Thanks in advance.

Edit:

I want to solve the actuall task now. But I can not do it myself. I observe a morphism $f:X\to Y$ in $(Set)_G$. Therefore a function with $f(xg)=f(x)g$ $\forall x,g$.

Now, I need to find a criterion, when $f(x)g=y$. I am not entirly show, what I actually have to do. Does the morphism $f$ has a stronger "role", than providing $f(xg)=f(x)g\,\forall x,g$? Do I have to check something first?

It is $f(xg)=f(x)g$ for all $x,g$. When is $f(x)=y$? That is the case, when $y\cdot g^{-1}\in Y$ for every $y,g^{-1}$, or is it? But the function $G\to Y, g\mapsto yg$ is a surjection, for every $y\in Y$, because $Y\times G\to Y$ is transitive.

We noted in the lecture, and I think that this task wants us to proof this, that every group actions is the disjoint union of transitive group actions. This should be the criterion, which is asked for. But what does it even mean for a group action, to be a disjoint union?

Answer 1

This is just a standard abuse of notation that you are surely familiar with in other contexts. A group is not just a set $G$, but actually a pair $(G,\cdot)$ where $\cdot$ is a binary operation on $G$, but we nevertheless frequently refer to "$G$" alone as the group. Similarly, a right $G$-set is really a set $X$ together with an action $X\times G\to X$, but we frequently refer to "$X$" alone as the $G$-set.

Answer 2

Morphisms of $G$-actions are the same as morphisms of $G$-sets. To clarify this, I need to introduce the necessary definitions.

By right $G$-action (on set $X$) we usually mean a function $\cdot\,\colon X\times G\to X$ such that $x\cdot(gh) = (x\cdot g)\cdot h$ and $x\cdot e_G = x$. Morphism between $G$-actions on sets $X$ and $Y$ are defined as functions $f\colon X\to Y$ such that $f(x\cdot_X g) = f(x)\cdot_Y g$. At this point I emphasized difference between actions on $X$ and $Y$, but this is usually left out because it is understood that $G$ acts differently on $X$ and $Y$.

By $G$-set we usually mean a set $X$ equipped with an action of $G$. So, morphism of $G$-sets $(X,\cdot_X)$ and $(Y,\cdot_Y)$ is a function $f\colon X\to Y$ such that $f(x\cdot_X g) = f(x)\cdot_Y g$. Again I emphasized difference between actions on $X$ and $Y$, but this is usually left out because it is understood that $G$ acts differently on $X$ and $Y$.

Now, why do I keep emphasizing difference between actions on $X$ and $Y$ when it is usually left out? Actually, because it is annoying. We never write "a linear map $A\colon (V,+_V,\cdot_V)\to (W,+_W,\cdot_W)$" because it is understood that $V$ and $W$ are vector spaces that have their own structure and $A$ is a map that respects it. It is enough to say that $A$ is linear. Same thing with $G$-actions - we won't say that $f$ is map between $G$-actions $X\times G\to X$ and $Y\times G\to Y$, but just $f\colon X\to Y$ where $G$-actions are understood. Actually, the whole "emphasizing" is probably set theoretically wrong and to make it precise would need more effort than the usual way.

I mentioned vector spaces because it might be an enlightening example since vector spaces are just abelian groups equipped with (left) action of a field, and linear maps are just group homomorphisms that respect the actions of the field.

Answer 3

The problem in question (a) is stated (by your professor I guess) in a terrible way... The given data of the problem is the G-sets $X$ and $Y$ and the points $x \in X$ and $y\in Y$ : it is then asked to find a property $\mathsf P$ depending on those data such that $\mathsf P$ is equivalent to the property "$\exists f \colon X \to Y$ morphism of $G$-sets such that $f(x)=y$". This is a very open question (for example you could choose $\mathsf P$ to be the sentence in quotes, but it would not be very satisfying).

To get a satisfying $\mathsf P$ (I mean an answer that will satisfy your professor), you should look at (b) and try to solve it before (a) and ask yourself what could be this $H$.

Hint. Try to find a property $\mathsf P$ that relates the elements of $G$ fixing $x$ and those fixing $y$.