TITLE: Understanding induced representations
QUESTION [28 upvotes]: Let $G$ be a group and $H$ be a subgroup. Let $\phi:H\rightarrow GL(V)$ be a representation of $H$. There are three constructions in Wikipedia, but I am not really convinced by these.
My question is: are there any good intuitive ways to understand induced representation $Ind_H^G(\phi)$?

REPLY [2 votes]: To begin we note that if we have $n\in\mathbb N$ and a rings homomorphism $\alpha:A\to B$, then naturally arises homomorphism $\alpha_n:M_n(A)\to M_n(B)$ such that $(X_{i,j})_{i,j}\mapsto(\alpha(X_{i,k}))_{i,j}$ and it gives a group homomorphism $\alpha_n:GL_n(A)\to GL_n(B)$ (since the invertible elements go into invertible under the action of homomorphism). 
Now let $G$ be a finite group, $H\leq G$. We can consider group algebra $F[G]$ as a right $F[H]$ module, which we denote $M$. We have a natural linear representation $r:G\to GL_{F[H]}(M)$ of $G$ over $F[H]$: $g\mapsto \varphi_g$, where $\varphi_g(x)=gx$. 
After these remarks we return to induced representations. Let $\rho:H\to GL_n(F)$ be a linear representation over a field $F$. Note that if $t=(t_1,\ldots,t_m)$ - left transversal of $H$ in $G$, then $t$ is a basis of $M$, hence we can go from $r$ to its version $r_t:G\to GL_m(F[H])$ in the basis $t$. We can extend $\rho$ by linearity to a homomorphism $\tilde\rho:F[H]\to M_n(F)$ of $F$-algebras. Then $\tilde\rho_m r_t:G\to GL_{mn}(F)$ is an induced representation.
This construction gives not only an induced representation, but also a transfer map $G\to H/N$, where $H'\leq N\leq H$. Indeed let $\nu:H\to H/N$ be a natural homomorphism and $\tilde\nu:F[H]\to F[H/N]$ be its extension by linearity to a homomorphism of $F$-algebras. Then $F[H/N]$ is a commutative ring, hence we have the right to consider determinants of the matrices with entries in $F[H/N]$. Then the map $g\mapsto \det(\tilde\nu_m r_t(g))$ is the transfer map. Since it is a composition of two homomorphisms, then transfer map is a homomorphism. This construction also allows for easy to prove independence of the transfer map from the choice of transversal - it's just a consequence of the fact that the conjugate matrices have the same determinants.