Induced hermitian module I've read about inducing representations of H modules to G modules, where H is a subgroup of G. If the H module has a hermitian form on it (hermitian with respect to the involution on k[H] sending h->h^{-1} ), does there exist a compatible hermitian form on the G module ?
Are there any references for this topic at a beginner's level ?
 A: I interpret the question as whether induced representation of unitarizable representation are unitarizable.
This is true, when $G$ is a locally compact group and $H$ a closed subgroup. It's due to Mackey. The construction involves a (quasi invariant) measure on $H \backslash G$, which behaves nicely under right translation. 
For the measure: The existence of the measure is shown in Folland "Harmonic analysis", but he focuses soon thereafter on the compact case. I do not know of any other textbook, which has reference  with proofs for this. 
For the induced representation: Barut and Raczka "The theory of group representation ... " do quite a good job in the general case, but omit many proofs and have a cumbersome notation (adopted from Mackey's paper). 
At a beginner's level? For compact groups, there are more reference. But I think that there a not so many sources, which explain Mackey's theory in general, and none at a beginners level. I suggest to learn a lot of finite/compact group stuff, before going to the more general settings, and consider induction and restriction as functor.
Attention: There are also non-unitarizable representations of $H$, which become unitarizabile after induction. The complementary series representations are examples.
For a finite group and there representation on characteristic zero fields, you can also write down the canonical inner product yourself, its
$$\langle f(g), h(g) \rangle_G =  \sum\limits_{H \backslash G} \langle f(g), h(g) \rangle_H,$$
and the well-behavedness in your question with respect to the action makes it well-defined. Perhaps you have learnt the tensor approach first, where this choice is not so obvious, but for function on $G$, this choice is obvious.
In characteristic $p$ the inner product can of course be non-degenerate for obvious reasons.
