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It is well known that $\det(A \otimes B) = \det(A)^m \det(B)^n$ when $A$ and $B$ are square matrices of size $n$ and $m$ where $\otimes$ denotes the Kronecker product.

Question: Is there a similar formula for the permanent $per(A \otimes B) $ (instead of the determinant)?

One motivation is to calculate the permanent of the character table of an abelian group since the permanent for cyclic groups is known and the character table of a direct product $G_1 \times G_2$ of groups is given as the Kronecker product of the character table of $G_1$ and $G_2$.

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    $\begingroup$ An equality of polynomial nature looks tricky in general: If $a$ and $b$ are integers greater than $1$, and $J_{a} $ and $J_{b}$ are the square all $1$'s matrices of respective sizes $a$ and $b$, then $J_{a},J_{b}$ and $J_{a} \otimes J_{b}$ have respective permanents $a!,b!$ and $(ab)!$. Note that there is always a prime greater then either of $a$ and $b$, but less than $ab$. $\endgroup$ Jul 2 '20 at 9:09
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In this paper:

R. A. Brualdi, Permanent of the direct product of matrices, Pacific J. Math. 16 (1966), 471482

(the Kronecker product is called here 'direct product') it is shown that, if $A$, $B$ are nonnegative matrices with order $m$, $n$, respectively, it holds true that: $$ \operatorname{per}(A \otimes B) \geq \operatorname{per}(A)^n \operatorname{per}(B)^m $$ where the equality holds iff $A$ or $B$ has at most one nonzero term in its permanent expression. Moreover, there exists a minimal number, denoted in the paper by $K_{m,n}$, such that: $$ \operatorname{per}(A \otimes B) \leq K_{m,n} \operatorname{per}(A)^n \operatorname{per}(B)^m $$ These numbers satisfy the inequality: $$ K_{m,n} \geq \frac{(mn)!}{(m!)^n (n!)^m} $$ and it is conjectured that we actually have an equality.

Furthermore, in the paper:

Marvin Marcus, Permanents ot direct products, Proc. Amer. Math. Soc. 17:226-231 (1966)

it is proven that if $A$, $B$ are positive semidefinite Hermitian square matrices of order $m$, $n$, respectively, then: $$ \operatorname{per}(A \otimes B) \geq \left (\frac{1}{n!} \right )^m \left (\frac{1}{m!} \right)^n \operatorname{per}(A)^n \operatorname{per}(B)^m $$

Equality holds iff at least one of $A$, $B$ has a zero row.

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