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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:

Give the dual $M^{\ast }={\mathrm{Hom}}_k(M,k)$ the contragredient action $(g\lambda )(m)=\lambda (g^{-1}m)$. If ${\delta }_h$ is the functional dual to the basis element $h\in G$, thenThus the map $h\mapsto {\delta }_h$ extends to a $kG$-isomorphismThis is also the left-module form of the fact that a group algebra is a symmetric algebra.

If $P$ is finitely generated and projective, it is a direct summand of $(kG{)}^r$. Dualizing makes $P^{\ast }$ a direct summand of $((kG{)}^{\ast }{)}^r\cong (kG{)}^r$, so $P^{\ast }$ is projective. The converse follows by dualizing again and using $P\cong P^{\ast \ast }$.

For any finite-dimensional algebra $A$, the module $A^{\ast }$ is injective becauseis exact. Since $kG\cong (kG{)}^{\ast }$, free $kG$-modules are injective, and so are their projective direct summands. Conversely, duality sends injectives to projectives, so every finite-dimensional injective is projective. Hence projective modules over a finite group algebra are injective.

Finally, let $P$ be indecomposable projective. It is also an indecomposable injective. Its nonzero socle contains a simple module $S$, and the injective hull $E(S)$ is a direct summand of $P$. Indecomposability forces $P=E(S)$. Since $S$ is essential in its injective hull, every simple submodule of $P$ equals $S$. Therefore