If we consider an induced representation $Ind\_{H}^{G}1\_{H}$ of finite groups, where $1\_{H}$ is the trivial caracter of $H$, how we decompose this representation into irreducible ? There is a decomposition $Ind\_{H}^{G}1\_{H}=V\oplus W$, where $V$ is formed of canstant functions on $G$ and $W$ is formed of functions $f:G\longrightarrow\mathbb{C}$, $f\in Ind\_{H}^{G}1\_{H}$, such taht $$\sum\_{x\in G/H}f(x)=0$$
but I don't know if $ W $ is irreducible and if it is not irreducible how decomposed there into irreducible ?


The question arises also for profinite groups : If $G$ is totaly disconnected compact group and $H$ is an open subgroup, how to decompose the representation $Ind\_{H}^{G}1\_{H}$ into irreducibles ?



Decomposition into irreducible of induced representations
Since it has not really been made completely explicit in the answers above, I point out that the specific question of whether your invariant subspace $W$ is irreducible or not is easier to check, at least in the case of finite groups. It rarely is. As is implicit in the earler answers, there is a formula due to Mackey for the norm (squared) of the induced character (and Mackey's formula works in more general contexts. In the case of finite groups, it is a double application of Frobenius reciprocity). We have $\langle {\rm Ind}\_{H}^{G}(1), {\rm Ind}\_{H}^{G}(1) \rangle = |T|$ , where $T$ is a full set of representatives for the distinct $(H,H)$-double cosets (so $G$ is the disjoint union $\cup\_{t \in T} HtH ).$ The space $W$ is irreducible if and only if this norm is $2,$ which happens if and only if there are just two $(H,H)$-double cosets. This in turn happens precisely when $G$ is doubly transitive in the usual action on the ( say right) cosets of $H.$


