# Corollary of Frobenius Reciprocity providing an exact condition for induced irreps of normal subgroups

I'm trying to understand a result from a course I'm taking, which is a corollary of Frobenius reciprocity.

It says that where $$\sigma$$ is an irrep of $$H \triangleleft G$$, then the induced representation $$\mathrm{Ind}_H^G \sigma$$ is irreducible if and onlty if for all $$g \in G \setminus H$$, the representation \begin{align} \sigma_g : H & \to \mathrm{GL}(W) \\ h & \mapsto \sigma(g^{-1} h g) \end{align} is not isomorphic to $$\sigma$$.

I don't see how this is derived from Frobenius Reciprocity. The text gives the hint that Frobenius reciprocity tells us that for $$\rho$$ an irrep of $$G$$, the multiplicity of $$\mathrm{Res}_H^G \rho$$ in $$\sigma$$ is precisely the multiplicity of $$\rho$$ in $$\mathrm{Ind}_H^G \sigma$$, but I don't see how this helps. Clearly that multiplicity must be $$0$$ for every $$\rho$$ but one, but it still seems like a big leap from that to the above result.

## 1 Answer

Okay, think I figured it out. (I'll be abusing notation and writing $$\sigma$$ and it's character identically.) $$\mathrm{Ind} \sigma$$ will be irreducible if and only if \begin{align} 1 & = \left< \mathrm{Ind}\sigma, \mathrm{Ind}\sigma\right>_G \\ & = \left< \sigma, \mathrm{ResInd} \sigma \right>_H \\ & = \frac1{\# H} \sum_{h\in H} \sigma(h) \overline{\mathrm{ResInd} \sigma (h)} \\ & = \frac1{(\# H)^2} \sum_{h\in H \\ g \in G} \sigma (h) \overline\sigma(g^{-1} h g) \\ & = \frac1{\# H} \sum_{g\in G} \left<\sigma(h),\sigma_g(h) \right> \tag{1} \\ & = \frac1{\# H} \left(\sum_{g\in H} \left<\sigma(h),\sigma_g(h) \right> + \sum_{g\in G \setminus H} \left<\sigma(h),\sigma_g(h) \right> \right) \\ & = 1 + \sum_{g\in G \setminus H} \left<\sigma(h),\sigma_g(h) \right> \tag{2} \\ \iff & \sum_{g\in G \setminus H} \left<\sigma(h),\sigma_g(h) \right> =0 \end{align} where (1) is by the formula for character of an induced representation and (2) is by characters being invariant on conjugate elements (we can only apply this when $$g \in H$$ as $$\sigma$$ is only a representation of $$H$$.