Definition: A module $M$ is called linearly compact if for a family of cosets $\{x_{i}+M_{i}\}_{\triangle}$, $x_{i}\in M$, $\triangle$ is a directed set, and submodules $M_{i}\subset M$ (with $M/M_{i}$ finitely cogenerated), the intersection of any finitely many of these cosets is not empty, then also $\bigcap_{\triangle} (x_{i}+M_{i})\neq 0$ (Wisbauer, Foundation of Module and Ring Theory, 29.7).
Properties of Linearly Compact Modules: Let $N$ be a submodule of the $R$-module $M$.
- Assume $N$ to be linearly compact and $\{M_{i}\}_{\triangle}$ to be an inverse family of submodules of $M$. Then $N+\bigcap_{\triangle} M_{i}=\bigcap_{\triangle}(N+M_{i})$.
- $M$ is linearly compact module if and only if $N$ and $M/N$ are linearly compact.
- Assume $M$ to be linearly compact. Then: (i). there is no non-trivial decomposition of $M$ as an infinite direct sum; (ii) $M/Rad\text{ }M$ is semisimple and finitely generated; (iii) every finitely generated module in $\sigma[M]$ is linearly compact (Wisbauer, Foundation of Module and Ring Theory, 29.8).
My point is to prove point 3(iii). In the book, the clue is as a consequence of (2). If I have $X\in \sigma[M]$ and $X$ is finitely generated, then there is an $M$-generated module $K$ with $X\subset K$ and $\{U_{i}\}_{i=1}^{n}$ such that $\bigoplus_{i=1}^{n}U_{i}\rightarrow X\rightarrow 0$. Since $K$ is an $M$-generated module, then $M^{(\Lambda)}\rightarrow K\rightarrow 0$. How to connect $X$ as a submodule or quotient of $M$ in this proof? Thank you in advance.