= Solution
For a finite-dimensional module over $A$, the <radical of a module> satisfies $J(M)=J(A)M$. Hence
$$
J(U\oplus V)=J(A)(U\oplus V)=J(U)\oplus J(V),
$$
and induction gives $J^r(U\oplus V)=J^r(U)\oplus J^r(V)$ for every $r$. A simple submodule of a direct sum projects into semisimple submodules of each summand, and equivalently
$$
\operatorname{Soc}(M)=\{m:J(A)m=0\}.
$$
Thus the same argument, or induction through the defining quotients, gives
$$
\boxed{\operatorname{Soc}^r(U\oplus V)
=\operatorname{Soc}^r(U)\oplus\operatorname{Soc}^r(V).}
$$
This is <radical and socle series of a direct sum>.
Now let $G$ be a finite $p$-group and let $k$ have characteristic $p$. The <group algebra of a p-group in characteristic p is local>, with unique simple module $k$. The socle of its regular module is
$$
\operatorname{Soc}(kG)=(kG)^G
=k\sum_{g\in G}g,
$$
which is one-dimensional. If $kG=U\oplus V$ with both summands nonzero, finite length gives nonzero socles for $U$ and $V$, and the direct-sum identity would make $\operatorname{Soc}(kG)$ at least two-dimensional. Therefore
$$
\boxed{{}_{kG}kG\text{ is indecomposable}.}
$$
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