For a finite-dimensional algebra, the module radical is exactly the action of the Jacobson radical

If $M=0$, then $J(A)M=0$ and $\mathrm{rad}(M)=0$, so the claim is immediate. Assume from now on that $M\mathrm{\ne }0$. Let $N$ be a maximal submodule of $M$. Then $M\mathrm{/}N$ has no proper nonzero submodule, so it is simple by [L2]. Fix $m\in M$. If $m\in N$, then every $j\in J$ already satisfies $jm\in N$. If $m\mathrm{\notin }N$, then the nonzero coset $m+N$ generates the simple module $M\mathrm{/}N$, so the $A$-linear map ${\theta }_m:A\to M\mathrm{/}N,a\mapsto am+N$ is surjective. Its kernel $I_m:=\{a\in A:am\in N\}$ is therefore a maximal left ideal of $A$. By [F1], $J$ is contained in every maximal left ideal, so $J\subseteq I_m$ and hence $jm\in N$ for every $j\in J$. Thus $JM\subseteq N$. Since $N$ was arbitrary, $JM\subseteq \mathrm{rad}(M)$.

Consider the quotient module $M\mathrm{/}JM$. Because $J$ acts trivially on it, $M\mathrm{/}JM$ is naturally a module over the semisimple ring $A\mathrm{/}J$ from [L1]. Therefore [L3] makes $M\mathrm{/}JM$ semisimple, so the intersection of its maximal submodules is $0$. Maximal submodules of $M\mathrm{/}JM$ correspond exactly to maximal submodules of $M$ containing $JM$, and their intersection is $\mathrm{rad}(M)\mathrm{/}JM$. Hence $\mathrm{rad}(M)\mathrm{/}JM=0$.

Step 1.1 gives $JM\subseteq \mathrm{rad}(M)$ and step 2.1 gives the reverse inclusion. Therefore $\mathrm{rad}(M)=J(A)M$.