Local Representation Theory Alperin Ex 1.4. Let $A$ be a finite dimensional $k$-algebra where $k$ is an algebraically closed field. Let $U$ be an $A$-module of Loewy length $s$. We wish to show $$rad^i(U) \subseteq soc^{s-i}(U)$$ for $0 \leq i \leq s$ # Attempt Clearly, $rad^0(U) \subseteq soc^s(U) = A$. Now, we wish to show the $i-1$ case implies the $i$ case. Suppose $$rad^{i-1}(U) \subseteq soc^{s-(i-1)}$$ We have $$rad^{i}(U) = rad(U) soc^{s-(i-1)}$$ but $$rad(U)soc^{s-(i-1)}(U) \subseteq soc^{s-(i-1)-1}(U)=soc^{s-i}$$ and so the result is complete.

Here's an alternative proof. Let $J$ be the Jacobson radical of $A$. Since $A$ is finite dimensional, we know that $J$ is nilpotent and that $A/J$ is semisimple. Thus, for any (left) $A$-module $M$ we have $$ \mathrm{rad}^i(M) = J^iM \quad\textrm{and}\quad \mathrm{soc}^i(M) = \\{ m\in M : J^im=0 \\}. $$ Now, if $M$ has Loewy length $s$, then $J^sM=0$, so $J^{s-i}\mathrm{rad}^i(M)=0$, so $\mathrm{rad}^i(M)\subseteq\mathrm{soc}^{s-i}(M)$.