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Question: Let $\chi$ and $\psi$ be characters of $G$. Express $\det(\chi\psi)$ in terms of $\det\chi$ and $\det\psi$.

This is problem 4.1 from Isaacs' book. My thought was to let $X$ be a representation affording $\chi$, let $Y$ be a representation affording $\psi$, then we have that $X \otimes Y$ affords character $\chi\psi$ and is independent of the choice of bases from $X$ and $Y$. Now, products of characters are characters, so $\chi\psi$ is a character. Let $\{x_i:1\leq i\leq n\}$ and $\{y_r:1\leq r\leq m\}$ be bases for $X$ and $Y$. Let $g\in G$. Then, for $a_{ij},b_{rs}\in\mathbb{C}$, we have that $x_ig=\sum_{j=1}^na_{ij}x_j$ and $y_rg=\sum_{s=1}^mb_{rs}y_s$. Now, $(\chi\psi)(g)=\chi(g)\psi(g)$. I would love to say that since $\det X(g) = (\det\chi)(g)$ and $\det Y(g) = (\det\psi)(g)$ that the $\det(\chi\psi) = \det(X\otimes Y) = \det\chi \det\psi$.

But I feel like something is far off, because the problems hint says "it suffices to assume that $G$ is cyclic", but I never used that.

Nicky Hekster
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User7238
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    If $\chi$ is the character of $\rho:G\to GL(X)$, are we defining $\det\chi$ to be $\det\rho(g)$ as a function defined on $G$? If so, use https://math.stackexchange.com/questions/1316594/determinant-of-the-kronecker-product-of-two-matrices – anon Oct 06 '22 at 03:28
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    You find bases such that $X(g)$ and $Y(g)$ are diagonal matrices. Then this gives you a basis such that $(X \otimes Y)(g)$ is a diagonal matrix. The determinant of a diagonal matrix is just the product of the entries, so it should be easy to calculate. Note that the correct solution is not $\det(\chi) \det(\psi)$. – spin Oct 06 '22 at 07:04

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