y\rho_1^{-1}\gamma$ for all $x \in D_y$ and $y\in X\rho_{1}$. Since $\ker \rho_{1}=\ker \gamma$, the transformation $\rho_2$ is well defined and we have $\gamma = \rho_{1}\rho_{2}$. Moreover, $\rho_2 \in \left\langle \OP(X,Y),B \right\rangle$ by Case 1 (since $\ker \rho_2 \in \mathcal{R}_{k}$) and thus $\gamma =\rho_{1}\rho_{2}\in \left\langle \OP(X,Y),B \right\rangle$.

\[lenew\] $\left\langle \eta |_{Y}\right\rangle =\left\langle \left\{ \beta
|_{Y}:\beta \in \OP(X,Y)\right\} \right\rangle \cap \S(Y)$.

The inclusion $\left\langle \eta |_{Y}\right\rangle \subseteq \left\langle
\left\{ \beta |_{Y}:\beta \in \OP(X,Y)\right\} \right\rangle \cap \S(Y)$ is obviously. Let now $\beta \in \OP(X,Y)$ with $\beta |_{Y}\in \S(Y)$. Then there is $k\in \{1,\ldots ,m\}$ such that $$\beta =\left(
\begin{array}{cccccc}
A_{1} & \cdots  & A_{m-k+1} & A_{m-k} & \cdots  & A_{m} \\
a_{k} & \cdots  & a_{m} & a_{1} & \cdots  & a_{k-1}\end{array}\right)$$ with $\{A_{1},A_{2}<\cdots <A_{m}\} \in \mathcal{P}_{m} \cup \mathcal{Q}_{m}$ and $a_{i}\in A_i$ for $i\in \{1,\ldots ,m\}$ since $Y$ is a transversal of $\ker \beta $. Thus, $$\beta |_{Y}=\left(
\begin{array}{cccccc}
a_{1} & \cdots  & a_{m-k+1} & a_{m-k} & \cdots  & a_{m} \\
a_{k} & \cdots  & a_{m} & a_{1} & \cdots  & a_{k-1}\end{array}\right) =(\eta |_{Y})^{m-k+1}\in \left\langle \eta |_{Y}\right\rangle.$$ This shows that $\left\langle \left\{ \beta |_{Y}:\beta \in \OP(X,Y)\right\}
\right\rangle \cap \S(Y)\subseteq \{(\eta |_{Y})^{p}:p\in \mathbb{N}
\}=\left\langle \eta |_{Y}\right\rangle$.

The following lemmas give us necessary conditions for 