0))}\\
 &=&
 ((0,\dots , 0, f_{{e_j}^{(k)}+e_{r_{k}}^{(k)}}^{(k,k+1)} (t'_{k+1})-t'_{k+1}, 0,\dots, 0),(0,\dots ,0,s'_{k+1},0,\dots, 0))\in
 I\cap A_{k+1}
 \end{aligned}$$ for some $s'_{k+1}\in S_{k+1}$ and $$\begin{aligned}
\lefteqn{f_{{e_j}^{(k)}+e_{r_{k}}^{(k)}}^{(k,k+1)} (t'_{k+1})-t'_{k+1} }\\
&=&
( (f_{k+1} -{\operatorname{id}})(v_{1}), \dots ,  (f_{k+1} -{\operatorname{id}})(v_{j-1}),  (f^{2}_{k+1} -{\operatorname{id}})(v_{j}),  (f_{k+1} -{\operatorname{id}})(v_{j+1}),\\
&&
 \dots ,  (f_{k+1} -{\operatorname{id}})(v_{m_{k+1}})).\end{aligned}$$ Since $m_{k+1}\geq r_k>j$ and $f_{k+1}\neq{\operatorname{id}}$, there exists $v_{m_{k+1}}\in V_{k+1}$ such that $(f_{k+1}
-{\operatorname{id}})(v_{m_{k+1}})\neq 0$, thus Claim 1 yields that $$a_{k+1,r_{k+1}}\in I.$$ Therefore, applying the previous case, we get that $A_{k+2}\subseteq I$.

Now, combining the above two cases we clearly get that $A_{z}\subseteq I$, for all $z\in {\mathbb{Z}}/(n)$. So, $I=T\rtimes_{\circ}S$, as desired.

We now prove the main result of the paper, Theorem \[MainResult\].

[**(of Theorem \[MainResult\]).**]{} For each $z\in {\mathbb{Z}}/(n)$ let $V_z=({\mathbb{Z}}/(p_z))^{2(p_{z-1}-1)}$ and let $b_z\colon V_z\times V_z\longrightarrow {\mathbb{Z}}/(p_z)$ be the symmetric bilinear form with associated matrix $$\left(
\begin{array}{c|c}
0&I_{p_{z-1}-1}\\
\hline
I_{p_{z-1}-1}&0
\end{array}\right)$$ with respect the standard basis of $V_z$, where $I_m$ denotes the $m\times m$ identity matrix. Let $C_z$ be the companion matrix of the polynomial $x^{p_{z-1}-1}+x^{p_{z-1}-2}+\cdots +x+1\in
{\mathbb{Z}}/(p_z)[x]$, that is $$C_z=\left(
\begin{array}{ccccc}
0&0&\ldots&0&-1\\
1&0&\ldots&0&-1\\
0&\ddots&\ddots&\vdots&\vdots\\
\vdots&\ddots&\ddots&0&-1\\
0&\ldots&0&1&-1
\end{array}\right)\in GL_{p_{z-1}-1}({\mathbb{Z}}/(p_z)).$$ Let $f_z$ be the automorphism of $V_z$ with associated matrix $$\left(
\begin{array}{c|c}
C_z&0\\
\hline 0&(C_z^{-1})^t
\end{array}\right).$$ (For a matrix $A$, we denote by $A^t$ the transpose of $A$.) Note that $f_z$ 