a closed inclusion}\}$ and $\mathfrak R=\{R \subseteq \Bbbk[G]: R \text{ is a }
  \Bbbk\text{--subalgebra of } \Bbbk[G]\}$ and the maps:

1.  $\mathbb F: \mathfrak H \to \mathfrak R\,,\, \mathbb
        F(H):={}^H\Bbbk[G]=\{f \in \Bbbk[G]: x\cdot f=f, \forall x \in
        H\}$;

2.  $\mathbb S: \mathfrak R \to \mathfrak H\,,\, \mathbb S(R)=\operatorname{Stab}(R):=\{x  \in \Bbbk[G]: x\cdot r=r, \forall r \in R\}$.

In the above situation it is usual to write $\mathbb F(H)=H'$ and similarly, $\mathbb S(R)=R'$.

\[theo:connection\] In the above context, if we endow the sets $\mathfrak
  H\,,\,\mathfrak R$ with the order given by inclusion, the maps $\mathbb F, \mathbb S$ form an (order inverting) Galois connection. Moreover, the fixed subgroups for this connection, i.e. $\{H \in \mathfrak H: H''=H\}$ are the observable subgroups.

\[coro:firstproperties\] In the above situation one has that:

1.  For any $H \subseteq G$, $H''$ is observable in $G$;

2.  $H''=\bigcap \{K: H \subseteq K \subseteq G\,,\, K \text{
          observable}\}$;

3.  If $A$ is a commutative rational $G$–module algebra, and $H \subseteq G$ is a closed inclusion, then ${}^H A={}^{H''}A$.

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1.  For a Galois connection $H'''=H$ hence (1);

2.  In the above situation $H \subseteq K$ implies that $H''\subseteq K''=K$ and then $H'' \subseteq \bigcap\{K: H \subseteq K \subseteq 