Is this formula for induction always possible? I am beginning to learn representation and character theory. I worked out some proofs of the character table of the dihedral groups. But I have some issues understanding how deep is the method. Say $n$ its even and $G$ is the dihedral group of $2n$ elements. A way to find two nontrivial characters is to look at the subgroup generated by one of the generating rotations, say $H = \langle a^2 \rangle$ where $a$ is such a rotation. This group is normal in $G$ and $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Therefore we obtain four characters of $G$ by lifting the four of $G/H$, taking them to be trivial on $H$ and therefore being defined as the image of any representative of the class in $G/H$. 
I know that there is a more complicated notion of induction of representations, and here is my questions with respect to this: 


*

*is this what we call the induction of the (trivial ?) character from $H$ to $G$? from $G/H$ to $G$?

*is this procedure doable for any subgroup $H$? Or do we need specific properties such that $H$ normal? $G/H$ abelian? (I mean, to define the character as trivial on $G$ and on the value of a representative for a coset)

*In general, what is the strategy? Do we seek the largest possible $H$ so that $G/H$ is small and hence has simple enough characters?


Thanks in advance!
 A: 1,2) No, this is called "inflation".
The only thing you need is that $H$ is normal so that $G/H$ makes sense as a group, and it is simply defined as the composite $G\to G/H \to GL_n$. 
Induction is a bit more sophisticated (although it is still simple), and it allows to go from a representation of $H$ to $G$ here we went from $G/H$ to $G$) and it works for any subgroup of $G$ - but it doesn't preserve irreducibility in general. 
3) Well if you can find subgroups $H$ where you understand representations of $G/H$ fairly well, it is interesting to see the representations of $G$ you get this way; so that's a strategy, but often it can't work : if you look for instance at simple groups you have no interesting $H$, or groups like $S_n$ ($n\geq 5$) that have only $A_n$ as subgroups. But this is definitely one of the techniques, as if $V$ is irreducible as a $G/H$-representation, it is automatically irreducible as a $G$-representation, and isomorphism as $G/H$-representations is the same as isomorphism as $G$-representations so you don't have to worry about a thing.
