Induced Module by S3 Let $S_3=\{1,x,x^2,y,xy,x^{2}y | x^3=1, y^2=1, xy=yx^2\}$ be the permutation group $S_3$. Let $H=\{1,x,x^2 \}\le S_3$ .
If $L$ is the trivial one-dimensional H-module, then how to show that  ${Ind_{H}}^{G}(L)$ is a direct sum of 2 non-isomorphic one dimensional $S_3$ module.
 A: One of the nice things about the induction functor is that in many cases, we can ignore how exactly it is defined and just use one specific property of it, namely the one known as Frobenius reciprocity.
Frobenius reciprocity says that the irreducible constituents of $\rm{Ind}_H^G(L)$ are exactly those irreducible representations of $G$ which have $L$ as a constituent when restricted to $H$ (more precisely, the induction functor is adjoint to the restriction functor).
I will prove something more general and leave it as an exercise to adapt the proof to the specific case.
Let $G$ be a finite group and $H$ be a normal subgroup of $G$ such that $G/H$ is abelian. Let $1_H$ denote the trivial representation of $H$ (I will assume all representations are over $\mathbb{C}$).
Then $\rm{Ind}_H^G(1_H)$ is a sum of distinct 1-dimensional representations of $G$.
Proof: Note that since $G/H$ is abelian, we have $G'\leq H$ and since induction is transitive, we have that $\rm{Ind}_H^G(1_H)$ is a constituent of $\rm{Ind}_{G'}^G(1_{G'})$ so we can assume that $H = G'$.
Now Frobenius reciprocity tells us that the constituents of $\rm{Ind}_{G'}^G(1_{G'})$ are precisely those that have $1_{G'}$ as a constituent when restricted to $G'$. I claim that these are precisely the $1$-dimensional representations of $G$.
First, clearly the $1$-dimensional representations of $G$ have $G'$ in their kernel, which exactly means that restricting to $G'$ gives $1_{G'}$ (so this is clearly a constituent).
On the other hand, we know that the dimension of $\rm{Ind}_{G'}^G(1_{G'})$ is the index of $G'$ in $G$, which is also equal to the number of distinct $1$-dimensional representations of $G$, so since all of these are constituents, there cannot be any others, and the proof is complete.
(Part of the above could probably have been made a bit shorter by using that $G'$ is normal and using a bit of Clifford theory, but I wanted to make it as elementary as possible).
A: I assume you are doing tensor induction.  Then one useful formula to know is that if $k$ is the trivial $H$-module then $\operatorname{Ind}_G^H k = k[G/H]$, where by $k[G/H]$ I mean the module corresponding to the permutation representation on cosets of $H$.  So it's the same construction as the group ring $k[G/H]$ only in general you only get a $k[G]$-module instead of a ring (in this case you get a ring because $H$ is normal so $G/H$ is a group).
The basis for $k[G/H]$ is $\{1, y\}$.  The $G$-module structure is that $x$ acts as the identity and $y$ swaps the two basis elements.  Now to show it decomposes consider the basis $\{1 + y, 1 - y\}$.
