The radical and socle series of a module and its dual I am study Peter Webb's book "A course in finite group representation theory", and stuck on ex.7  of chapter 6. The exercise is about the relationship between the socle series and radical series of a module and those of the dual module which is stated as below:
Let $k$ be a field, $G$ be a finite group, $U$ be a finitely generated $kG$-module and $U^*$ be its dual. Write ${\rm Soc}(U)$ as the socle of $kG$-module U and ${\rm Rad}(U)$ as the radical of $U$. Show that for each $n$
$$ {\rm Soc}^n(U^*)=\{f\in U^*|~f({\rm Rad}^n(U))=0\}$$
and
$$ {\rm Rad}^n(U^*)=\{f\in U^*|~f({\rm Soc}^n(U))=0\}.$$ 
I already know that
(i) for any $kG$-module M, we have 
$${\rm Rad}^n(M)=({\rm Rad}(kG))^nM \quad \mbox{and}\quad{\rm Soc}^n(M)=\{x\in M| ({\rm Rad}(kG))^nx=0\}.$$
(ii) We may define a nondegenerate bilinear form $<~,~>:U^*\times U\rightarrow k$ by 
 $<f,u>=f(u)$ for all $f\in U^*, u\in U$. Then what we want to prove may be rephrased as ${\rm Soc}^n(U^*)=({\rm Rad}^n(U))^{\perp}$ and ${\rm Rad}^n(U^*)=({\rm Soc}^n(U))^{\perp}$.
Not sure if these help. 
 A: Let $V$ be a $kG$-module, and let $W$ be a submodule of $V$. The submodule of $V^*$ consisting of those maps which vanish on $W$ is isomorphic to $(V/W)^*$. (Moreover, $V^*/(V/W)^*\cong W^*$).
So, to rephrase your problem: Show that $\DeclareMathOperator{Soc}{Soc}\DeclareMathOperator{Rad}{Rad}\Soc^n(U^*)=(U/\Rad^n(U))^*$ and $\Rad^n(U^*) = (U/\Soc^n(U))^*$.

We'll need the following:

$U$ is simple if and only if $U^*$ is simple. $U$ is semisimple if and only if $U^*$ is semisimple.

Proof: If $U$ is not simple with nonzero proper submodule $V$, then $(U/V)^*$ is a nonzero submodule of $U^*$, and is a proper submodule of $U^*$ since the quotient is isomorphic to $V^*\neq 0$, so $U^*$ is not simple.  Conversely, by what we have just shown, if $U^*$ is not simple then neither is $U\cong U^{**}$. 
    Since "*" distributes over direct sums, the dual of a direct sum of simple modules is again a direct sum of simple modules, so if $U$ is semisimple, then so is $U^*$. Finally, if $U^*$ is semisimple, then so is $U^{**}\cong U$. QED
So in the case $n=1$, $U/\Rad(U)$ is the largest semisimple quotient of $U$, so  $(U/\Rad(U))^*$ is the largest semisimple submodule of $U^*$, i.e. $(U/\Rad(U))^* = \Soc(U^*)$. (Edit: I had mistakes in my proof for the other one, so I took it out. But it "should be" similar.)
Without thinking too much more about the general case for $n$, it feels like this argument can either be generalized in some way, or maybe what I've shown is the base step in an induction argument. I'll leave that to you, but hopefully this gets you somewhere. 
