Galois Connection Between Posets

Question

I have the next doubt about this problem:

In a Galois Connection between posets, show that the subset $\{p\mid p=RLp\}$ of $P$ is equal $\{p\mid p=Rq \; for\; some \; q\}$ and give a bijection from this set to the subset $\{q\mid q=LRq\}$ of $Q$. What are these sets in the case of a group of automorphism of a field? and Does this generalize to an arbitrary adjunction?

I have problem with the question, for the first one:

Let it be $G$ a group of automorphism of a field $U$, $P=P(U)$ the set of all subsets $X\subset U$ ordered by inclusion, while $Q=P(G)$ the set of all subsets $S\subset G$ also ordered by inclusion ($S\leq S'$ if and only if $S\subset S'$ ).

Let $LX=\{\sigma\mid x\in X \; implies\; \sigma(x)=x\}$, $RS=\{x\mid \sigma\in S \; implies\; \sigma(x)=x \}$, in other words $LX$ is the subgroup of $G$ which fixes all points $x\in X$ and $RS$ is the set of fixed points of the automorphisms of $S$

I do not see if the sets $\{p\mid p=RLp\}$,$\{p\mid p=Rq \; for\; some \; q\}$, and $\{q\mid q=LRq\}$ are something that is known in this particular case.

For the second problem i am clueless.

Thank you for everything.

Answer 1

Hints:

1. From the Galois connection properties, deduce that $RLR\le R$ and $RLR\ge R$, hence $RLR=R$. Similarly, $LRL=L$.
2. If $p=Rq$, then $RLp=RLRq=Rq=p$.
3. The bijection is just given by $L$ on the fixed elements (its inverse by $R$).
4. $\{p\mid RLp=p\}$ contains subfields and $\{q\mid LRq=q\}$ contains subgroups.
   (When) do they contain all subfields/subgroups?