TITLE: Frobenius formula for the determinant
QUESTION [9 upvotes]: Is there a formula for the determinant of an induced representation, e.g. in the fashion of the Frobenius character formula.
I would hope for something:

$$ det \; Ind_H^G \rho(g) = (-1)^\alpha \prod\limits_{\gamma, \gamma_1 \in G/H \atop \gamma^{-1}g\gamma_1 \in H} det \rho(\gamma^{-1}g\gamma_1).$$ 

If you have a reference or a quick proof, that would be most helpful.

REPLY [8 votes]: The answer can be formulated in compact form using the usual transfer map $V_H^G\colon G\to H/H'$ or the (not-so-well-known) construction of tensor induction: Namely, 

$$ \det( { \rm Ind}_H^G\; \rho ) 
               = ( {\rm sign}_{[G:H]})^{\rho(1)} ((\det\rho)\circ V_H^G) 
               =( { \rm sign }_{ [G:H] })^{\rho(1)} (\det \rho)^{\otimes G}, $$

where ${\rm sign}_{[G:H]}$ is the permutation sign character of $G$ on the cosets of $H$. An exposition of tensor induction and a proof of this formula is contained in Curtis and Reiner, Methods of Representation Theory I (see Proposition 13.15). It seems this formula is due to Gallagher (MR0185017).
Of course, when evaluating at $g\in G$, this yields the last formula in Geoff Robinson's answer.