Let $R$ be a commutative unital ring and $n\in\Bbb{N}$, $n\geq 2$. A matrix $ \sigma = (\sigma_{ij})$, $1\leq{i, j} \leq{n}$, of additive subgroups $\sigma_{ij}$ of the ring $R$ is called a net or carpet over the ring $R$ of order $n$ if $\sigma_{ir} \sigma_{rj} \subseteq{\sigma_{ij}}$ for all $i$, $r$, $j$. A net without diagonal is said to be an elementary net or elementary carpet. Suppose that $n\geq 3$. Consider a matrix $\omega = (\omega_{ij})$ of additive subgroups $\omega_{ij}$ of the ring $R$, where $\omega_{ij}$, $i\neq{j}$, is defined by the rule: $ \omega_{ij} = \sum\limits_{k=1}^{n}\sigma_{ik}\sigma_{kj}$, $k\neq i,j$. The set $\omega = (\omega_{ij})$ of elementary subgroups $\omega_{ij}$ of the ring $R$ is an elementary net, which is called elementary derived net. The diagonal of the derived net $\omega$ is defined by the formula $\omega_{ii}=\sum\limits_{k\neq s}\sigma_{ik}\sigma_{ks}\sigma_{si}$, $1\leq i\leq n$, where the sum is taken over all $1 \leq{k\neq{s}}\leq{n} $. The following result is proved. An elementary net $\sigma$ generates the derived net $\omega=(\omega_{ij}) $ and the net $\Omega=(\Omega_{ij})$, which is associated with the elementary group $E(\sigma)$, where $ \omega\subseteq \sigma \subseteq \Omega$, $\omega_{ir}\Omega_{rj} \subseteq \omega_{ij}$, $\Omega_{ir}\omega_{rj} \subseteq \omega_{ij}$ $(1\leq i, r, j\leq n)$. In particular, the matrix ring $ M(\omega)$ is a two-sided ideal of the ring $M(\Omega)$. For nets of order $n=3$ we establish a more precise result.