An identity about Jacobson Radical $\theta(J(R))=J(S)$ I am reading the proof of Corollary 13.2 from Isaacs' Algebra.
Okay, I get the intuitive idea but have couple of disconnections:
1-In the first paragraph it says:

Every simple right $S$-module can be seen as an $R$-module, and as
  such, it is simple

I am not sure that preserving of being simple is obvious, and i cant see it.
2-In the second paragraph it says:

Since every simple $R$-module is annihilated by $J(R)\supseteq \ker(\theta)$, each can be viewed as an $S$-module and so is
  annihilated by $J(S)$

I can see why every simple $R$-module can be viewed as $S$-module, but i cant put $\ker(\theta )$ in the business.
Would please help me to understand this proof? Thanks.
 A: In both cases, it sound like you haven't looked at the module actions closely enough.
If $A$ is a simple left $S$ module, we view it as an $R$ module with the action $ar:=a\theta(r)$. (Later I'll be recycling $A$ for use as an $R$ module.)

I am not sure that preserving of being simple is obvious

Perhaps this alternative characterization of simplicity is what you need. Saying $A_S$ is simple is equivalent to saying that $S$ acts transitively on its nonzero elements. (That is, if $a,b\in A\setminus\{0\}$, there exists $s\in S$ such that $as=b$.)  From the above module action, it's easy to see that $R$ acts transitively if $S$ does.

but i can't put ker(θ) in the business.

In this half, we're trying to take an $R$ module and make an $S$ module out of it using the mapping $\theta$. An optimist might expect that reversing the definition that we did before would work, that is, for $s=\theta(r)$, $as:=ar$. The problem is that this isn't well-defined unless $\ker(\theta)\subseteq ann(A)$. When this is satisfied, then the optimistic definition works. (I invite you to examine the details to prove well-definedness.)
In that half of the problem, the assumption that $A_R$ is simple puts $J(R)$ into the annihilator of $A$, but by the first part we have that $\ker(\theta)\subseteq J(R)$. That proves $\ker(\theta)\subseteq ann(A)$, allowing us to use the desired module action. 
