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Let $H$ be a subgroup of $G$. Let $\rho$ be an irr representation of $G$ induced from an irr representation $\theta$ of $H$. It is well known that $\rho$ is monomial if $\theta$ is monomial. Is it possible that $\rho$ is monomial but $\theta$ is not monomial?

In other words, can monomial representations induced from nonmonomial representations?

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According to Djokovic and Maizan, the Specht module $V_{(3, 1, 1)}$ of $S_5$ is monomial. This is a representation of dimension $6$, induced from a representation of dimension $3$ of $A_5$. Since $A_5$ has no subgroup of index $3$ (see here for example), this representation of $A_5$ cannot be monomial.

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  • $\begingroup$ Thanks for providing this nice example. Is there a general theory behind such phenomena? $\endgroup$ – Huangjun Zhu Oct 14 '14 at 15:33
  • $\begingroup$ None that I know of (besides Geoff's answer alongside this one). $\endgroup$ – Amritanshu Prasad Oct 15 '14 at 3:42
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As Amritanshu Prasad points out, the $6$-dimensional irreducible complex representation of $S_{5}$ is indeed monomial (with respect to a suitable basis), and thinking about how to prove this directly led me to a general observation: let $G$ be a finite group, and $\chi$ be a non-linear complex irreducible character of $G$ of minimal degree. Let $H$ be a proper subgroup of $G$ of index less than $2\chi(1).$ Let $\lambda$ be a linear character of $H$ of order $m$ ( that is, $\lambda^{m}$ is the trivial character, but no smaller positive integer power of $\lambda$ is trivial). Then if $m$ does not divide $[G:G^{\prime}],$ the induced character $\theta = {\rm Ind}_{H}^{G}(\lambda)$ is irreducible. For otherwise, (by Frobenius reciprocity), there must be a linear constituent $\mu$ of $\theta$ such that $\langle {\rm Res}^{G}_{H}(\mu),\lambda \rangle \neq 0.$ But then the order of $\mu$ must be divisible by $m,$ whereas the order of $\mu$ divides $[G:G^{\prime}],$ a contradiction.

To apply this result to $G = S_{5},$ take $H$ to be the normalizer of a Sylow $5$-subgroup. Then $[G:G^{\prime}] = 2,$ and $H$ has a linear character $\lambda$ of order $4,$ so the above result can be used to see that ${\rm Ind}_{H}^{G}(\lambda)$ is irreducible (for note that $S_{5}$ has no non-linear irreducible character of degree less than $4$).

Incidentally, taking $G = A_{5},$ and $K$ to be a Sylow $5$-normalizer of $G$, we see that $[G:G^{\prime}] = 1,$ that $[G:K] = 6$ and that $G$ has no non-linear irreducible character of degree less than $3.$ However, $K$ has a linear character of order $2$ such that ${\rm Ind}_{K}^{G}(\lambda)$ is not irreducible ( it is a sum of two irreducible characters of degree $3$). Hence the bound $[G:H] < 2 \chi(1)$ in the result above can't be sharpened in general.

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