G\,,\, K \text{ observable}\}$ and being $H''$ observable, the proof of this part is finished.

3.  It is clear that ${}^{H''}A \subseteq {}^H A$. Take now, $a \in {}^H A$ and let $V$ be a rational finite dimensional $G$–module that contains $a$. If we call $K=\{x \in G: x\cdot a = a\}$, then Theorem \[theo:follows\] guarantees that $K$ is observable. As $a$ is fixed by $H$ we deduce that $H \subseteq K$ and then $H'' \subseteq K$ and that means that $a$ is fixed by all the elements of $H''$.

[[@kn:hmbb Theorem 8],[@kn:gross14]]{}\[theo:follows\] Assume that $H \subseteq $ is a closed inclusion of affine algebraic groups and that there is a finite dimensional rational $M \in {{}_{G}\mathcal M}$ with the property that there exists $m_0 \in M$ such that $H=G_{m_0}$. Then $H$ is observable in $G$.

Take an arbitray $\alpha \in M^*$. The element $\alpha|m_0$ satisfies the following equivariance condition for all $x \in G$ $x
\cdot (\alpha|m_0)= \alpha| (x \cdot m_0)$. Then, $\alpha|m_0 \in {}^H
\Bbbk[G]$. Assume now that $z \in G$ is such that $z \cdot f = f$ for all $f \in {}^H\Bbbk[G]$. Then for all $\alpha$ we have that $z \cdot
(\alpha|m_0)=\alpha|m_0$ and this implies that 