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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.

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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 $H$.

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