Some Preliminary Definitions and Properties:
Actions of a group $G$ on sets $X$ and $Y$ are equivalent if the corresponding action of $G$ on maps from $X$ to $Y$ fixes some bijection. In this case, we write $X \sim Y$.
An action is said to be transitive if it has only one orbit.
Given an action of a group $G$ on a set $X$,
the stabilizer, $G_x$, of $x$ is defined by
$$ G_x = \{ g \in G : gx = x \}; $$
and the orbit, $Gx$, of $x$ is defined by
$$ Gx = \{ gx : g \in G \}. $$
Properties:
(i) $Gx \subset X$ and $G_x \le G$;
(ii) there exists a bijection from $Gx$ to the coset space $G/G_x$; and
(iii) if $G$ is finite, then $|Gx|=|G/G_x|$.
The Statement of the Problem:
Prove the following:
(a) For each subgroup $H$ of a group $G$, the action $G$ by left translation on the left coset space $G/H$ is transitive.
(b) Each transitive action of a group $G$ on a set $X$ is equivalent to the action $G$ by left translation on the coset space $G/G_x$, for each $x \in X$.
(c) For subgroups $H$ and $K$ of $G$, the left translation actions of $G$ on $G/H$ and $G/K$ are equivalent if and only if $H$ and $K$ are conjugate.
Where I Am:
For part (a): The action of $G$ by left translation on the left coset space $G/H$ has only one orbit and is, thus, transitive. (i.e., since $gH=g \cdot H$, the orbit of $H \in G/H$ is the whole coset space.) [I pretty much understand this. It was mostly the result of piecing together definitions and properties, but doesn't make too much intuitive sense to me.]
For part (b): Since $Gx \subset X$ and we are considering transitive actions of $G$ on $X$ (i.e., actions with only one orbit), in this case, $Gx = X$ [I'm not actually sure if this is true, but it seems correct. I would like to be able to prove it, but can't seem to figure out how]. We know that there exists a bijection $f: Gx \to G/G_x$, so it must also be the case that $f$ maps from $X$ to $G/G_x$ bijectively. [Here, I need to prove that the action of $G$ on $f$ fixes $f$, which I can't figure out how to do.] Since the action of $G$ on $f$ fixes $f$, we have that $X \sim G/G_x$, as desired.
For part (c): I have made the least amount of progress on this one. To show that $H$ and $K$ are conjugate, I assume that I need to show that $K = gHg^{-1}$, for some $g \in G$, and $H = gKg^{-1}$, for some $g \in G$. I can see how this'd be the case if these subgroups were $normal$ in $G$, but I can't seem to get it from the assumptions given. Obviously, this statement is biconditional, so I have to prove the other direction, as well; but I'd be happy to get at least One Direction$^{\text{TM}}$.
Best Answer
Since I refered to this thread while answering another, I might as well answer this one, albeit being more than two years late. First, yes, if $X$ is a left $G$-set, then $G$ acts on $X$ transitively if and only if $G\cdot x=X$ for at least one (whence for any) $x\in X$.
For Part (a), you got it correctly. For Part (b), for a fixed $x\in X$, we define a function $f:X\to G/G_x$ as $f(y)=gG_x$, if $y\in X$ and $g\in G$ satisfy $y=g\cdot x$ (since $G$ acts on $X$ transitively, such an element $g\in G$ exists for every $y\in X$). Here, $G_x$ is the stabilizer of $x$, namely $G_x:=\big\{g\in G\,\big|\,g\cdot x=x\big\}$. We need to prove that $f$ is well defined. Suppose that $y=g\cdot x$ and $y=h\cdot x$ for $g,h\in G$. We have to prove that $gG_x=hG_x$. This is easy, since $$(h^{-1}g)\cdot x=h^{-1}\cdot(g\cdot x)=h^{-1}\cdot y=h^{-1}\cdot(h\cdot x)=(h^{-1}h)\cdot x=1_G\cdot x=x\,,$$ whence $h^{-1}g\in G_x$, and so $gG_x=hG_x$. Injectivity of $f$ is clear: if $f(y)=f(z)$ for some $y=g\cdot x$ and $z=h\cdot x$ with $g,h\in G$, then $g G_x=h G_x$, or $h^{-1}g\in G_x$, whence $$y=g\cdot x=(hh^{-1}g)\cdot x=h\cdot\big((h^{-1}g)\cdot x\big)=h\cdot x=z\,.$$ Surjectivity of $f$ is also immediate: for $g\in G$, we have $f(g\cdot x)=g G_x$.
For Part (c), we assume first that there exists an isomorphism of left $G$-sets $\phi:G/H\to G/K$. Suppose that $\phi(H)=kK$ for some $k\in G$. We know that $h\in H$ implies that $hH=H$, so $$kK=\phi(H)=\phi(hH)=\phi(h\cdot H)=h\cdot\phi(H)=hkK\,.$$ This proves that $k^{-1}hk\in K$ for every $h\in H$. That is, $k^{-1}Hk\subseteq K$. On the other hand, $$\phi(k^{-1}H)=\phi(k^{-1}\cdot H)=k^{-1}\cdot\phi(H)=k^{-1}\cdot (kK)=K\,.$$ If $t\in K$, then $tK=K$, so $$\phi(k^{-1}H)=K=tK=t\cdot K=t\cdot \phi(k^{-1}H)=\phi\big(t\cdot (k^{-1}H\big)=\phi(tk^{-1}H)\,.$$ Since $\phi$ is injective, $k^{-1}H=tk^{-1}H$, or $H=ktk^{-1}H$, which means $kKk^{-1}\subseteq H$, or equivalently $K\subseteq k^{-1}Hk$. That is, $K=k^{-1}Hk$ is a conjugate of $H$. Conversely, if $K=k^{-1}Hk$ for some $k\in G$, then we can define an isomorphism of left $G$-sets $\phi:G/H\to G/K$ by sending $gH\mapsto gkK$ for all $g\in G$. It is not difficult to show that $\phi$ is well defined, injective, surjective, and compatible with the left $G$-actions on $G/H$ and on $G/K$.
In addition, if $X$ and $Y$ are transitive left $G$-sets, then $X$ and $Y$ are isomorphic as left $G$-sets if and only if, for some $x\in X$ and $y\in Y$, the stabilizing subgroups $G_x$ and $G_y$ are conjugate subgroups of $G$. This follows immediately from Part (b) and Part (c) of this problem.
P.S. To avoid possible stupid problems with the empty set, all sets that appear in this answer are inherently assumed to be nonempty.