In representation theory, we consider the restriction functor for any group $G$ and subgroup $H$. This is:

$Res_H^G : Rep(G) \rightarrow Rep(H)$

This gives a representation of $H$

The Induced case lets us look in the other direction, going from subgroup representations to the groups representation

$Ind_H^G : Rep(H) \rightarrow Rep(G)$

How does this relate to Frobenius Reciprocity?

For example, I have the following question that I would like to be able to solve

Let $G$ be a finite group and $H$ be a subgroup. Let $1_H$ be the trivial representation of $H$. Show that the trivial representation $1_G$ of $G$ occurs exactly once in the induced representation $Ind_H^G1_H$

My instinct would be to assume $1_G$ appears more than once in the induced rep, and arrive at a contradiction. I have not yet been sucessful in this and would very much appreciate your help

  • 1
    $\begingroup$ Since $1_H$ is the restriction of $1_G$ to $H$ and $1_H$ clearly occurs exactly once as a constituent of $1_H$, the result follows immediately from the Frobenius Reciprocity Theorem, $1=(1_H,1_H)=((1_G)_H,1_H) = (1_G,1_H^G)$. $\endgroup$
    – Derek Holt
    Commented Apr 14, 2016 at 14:19

1 Answer 1


Frobenius Reciprocity says that induction and restriction are adjoint functors. If you aren't comfortable with that language, what this means is that if $V$ is a representation of $H$ and $W$ is a representation of $G$ then there is an isomorphism:

$$Hom_G(Ind_H^G (V), W) \cong Hom_H(V, Res^G_H (W)) $$

Really adjunction gives something stronger, a sort of consistent way to choose such an isomorphism for all such pairs $V$ and $W$, but for a lot of purposes this statement is what we really care about.

Applying this to your question, where $V = 1_H$ and $W = 1_G$, we get:

$$Hom_G(Ind_H^G (1_H), 1_G) \cong Hom_H(1_H, Res^G_H (1_G)) $$

The point is that while the left hand side involves a term $Ind_H^G (1_H)$ we don't quite understand, the left hand side is easy since clearly $Res^G_H (1_G) = 1_H$. So this just becomes

$$Hom_G(Ind_H^G (1_H), 1_G) \cong Hom_H(1_H, 1_H) \cong \mathbb{C}$$

So this tells us that $Hom_G(Ind_H^G (1_H), 1_G)$ is one dimensional, which exactly means that there is a single copy of the trivial representation of $G$ in $Ind_H^G (1_H)$, since if there were two copies we could project onto each one and scale them independently giving us a(n at least) two dimensional Hom-space.


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