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The head, or top, of a finite-length module $M$ is its largest semisimple quotient. It is $M/J(M)$, where $J(M)$ is the radical of a module.

For a finite-dimensional module over $A$, the radical of a module satisfies $J(M)=J(A)M$. Henceand 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 equivalentlyThus the same argument, or induction through the defining quotients, givesThis 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 iswhich 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 $\mathrm{Soc}(kG)$ at least two-dimensional. Therefore

The radical of a module is the smallest submodule $N\subseteq M$ for which $M/N$ is a semisimple module. Consequently, ifhas semisimple successive quotients, then $J(M_i)\subseteq M_{i+1}$. Induction givesso the radical series of a module descends at least as fast as every such series.

Dually, the socle is the largest semisimple submodule. Ifhas semisimple successive quotients, induction in $M/N_i$ givesso the socle series of a module ascends at least as fast as every such series.

Both series terminate because $M$ has finite composition length. More precisely,Thus $J^r(M)=0$ exactly when $J(A{)}^r$ annihilates all of $M$, which is exactly when ${\mathrm{Soc}}^r(M)=M$. The two least terminating indices therefore coincide: