Prove that a map is one-to-one if $G$ is finite group and $H$ and $K$ are its subgroups Okay, so I've been struggling with one problem lately. I know it must be something very obvious and the solution is right in front of me, but I just can't see it.
Let's get to the problem:
Let $G$ be a finite group.
If $H$ and $K$ are subgroups of $G$, then the map $f \mapsto f(H)$ is one-to-one correspondence between the set of $G$-set homomorphisms from $G/H$ to $G/K$ and the set of cosets $xK \in G/K$ such that $H \subseteq {}^{x}K$, where ${}^{x}K = xKx^{-1}, x \in G$. In particular. the $G$-sets $G/H$ and $G/K$ are isomorphic if and only if $H$ and $K$ are conjugate in $G$.
So, the first part seems to be very obvious. Let $\psi$ denote the map. Then we have to prove that $\psi(f) = \psi(g) \implies f = g$, where $f, g \in Hom(G/H, G/K)$. So let's start with that indeed. $\psi(f) = \psi(g) \iff f(H) = g(H)$. What should be the next step? My first thought was $\forall_{h \in H} f(h) = g(h)$, but that's clearly false.
Speaking of the second part, I am also stuck, but let me show you my thoughts on "$\impliedby$". Groups are conjugate, so $K = gHg^{-1}, g \in G$. From that we acquire that $gH = Kg$. Now we need to define a map that is an isomorphism of $G$-sets (bijective $G$-equivariant map). I suggest $tK \mapsto tKg = tgH$. That map is indeed $G$-equivariant, because: $\phi(g' \cdot tK) = g'tKg = g' \cdot tKg = g' \cdot \phi(tk), g' \in G$. But what about proving injectiveness and surjectiveness? How the conditions should look like?
Thank you in advance for any help.
 A: Well, $f\in\hom_G(G/H,\,G/K)$ is not defined on elements of $G$ (or $H$), so writing $f(h)=g(h)$ makes no sense.
The given part of the corollary is correct.
Observe that, if $f$ is $G$-equivariant $G/H\to G/K$, then $f(H)$  already determines it all, because we must have
$$f(x\cdot H)\ =\ x\cdot f(H)$$
so, if $f(H)=f'(H)$ for some other $f'$, then they must agree: $f(xH)=f'(xH)$ for all $xH\in G/H$, hence $f=f'$.
Let $S:=\{x\in G:\,H\subseteq xKx^{-1}\}$. Then we still need to prove $f(H)=xK$ with $x\in S$, and that the map $\psi$ is surjective to $\{xK:x\in S\}$.
For an arbitrary coset $xK\in G/K$, its stabilizer subgroup is just ${}^xK\subseteq G$.
Since $f(H)\in G/K$, there is an $x\in G$ such that $f(H)=xK$, and because $f$ is equivariant, for any $h\in H$,
$$xK=f(H)=f(hH)=hxK\,\implies\,x^{-1}hx\in K\,\iff\,h\in {}^xK\,.$$
To prove surjectivity, let $x\in S$, and show that $f:=gH\mapsto gxK$ is a well-defined $G$-equivariant map $G/H\to G/K$.
For the corollary, let $f:G/H\to G/K$ be a $G$-set isomorphism with inverse $f^{-1}:G/K\to G/H$.
The statement says that $f(H)=xK$ for some $x\in S$, so that $H\subseteq {}^xK$, and the same is true for $f^{-1}$, hence $K\subseteq {}^yH$ for some $y$.
So using finiteness, both these must be satisfied with equalities.
