Identifying an action on an arbitrary set with an action on corresponding factor group

Question

My textbook (Szekeres's A Course in Mathematical Physics) asks me the following set of questions (NB I am self-studying):

If $H$ is any subgroup of a group $G$, define the action of $G$ on $G/H$ by $\psi: G \to$ Sym($G/H$) with $\psi(g)(g'H) = gg'H$.

a) Show that this is always a transitive action of $G$ on $G/H$

b) Let $G$ have a transitive left action on a set $X$, and set $H = G_x$ to be the isotropy (stabilizer) group of any $x \in X$. Show that the map $i: G/H \to X$ defined by $i(gH) = gx$ is well-defined and bijective. [I note here that the transitive action alluded to is tacit in the notation $gx$. Call it $\varphi$ as required.]

c) Show that the left action of $G$ on $X$ can be identified with the action of $G$ on $G/H$ defined in (a).

Now I've done (a) and (b), but am a bit confused by (c). I assume it's alluding to there being some commutative diagram-type argument that I can make, saying something like "for every transitive action $\varphi$ of $G$ on $X$, we can think of it instead as $\varphi = i \circ \psi$, but I'm not sure if I'm missing something big. I really can't quite see how to start since the "identified with" is so vague here.

Answer 1

Typically, when a mathematics problem asks you to "identify" a mathematical object $A$ with a mathematical object $B$, the request is quite precise: write down a formula for a bijection $f : A \to B$ (or the other direction if you prefer), and prove that this bijection "preserves" the structure.

In your case, assuming that you have already solved part (b), you already have in your hands a formula for a bijection $i : G/H \to X$. The remaining step in part (c) is to show that this formula "preserves the action". This is still somewhat vague, but only because it applies across many, many different kinds of mathematical structures. In the case that the structure is a group action, what you are asked to verify is this:

For each $g \in G$ and each $g'H \in G/H$, $g(i(g'H)) = i(g(g'H))$,

which is precisely the "commutative diagram-type argument" that you intuited.

Answer 2

Let $X$ and $Y$ be two sets that a group $G$ acts on, in other words, two $G$-sets.

As $G$-sets $X$ and $Y$ are said to be isomorphic, if there exists a bijection $f: X \rightarrow Y$ such that $f(gx) = gf(x)$ for all $g \in G$ and $x \in X$.

Basically what this means is that the action of $G$ on $X$ and $Y$ is exactly the same, and you can use the map $f$ to identify $X$ and $Y$ as $G$-sets.

In your case it should be clear that the map $i: G/H \rightarrow X$ defined in (b) is an isomorphism of $G$-sets (once you have proven it is a well-defined bijection), so using the map $i$ we can identify $G/H$ and $X$ as $G$-sets.