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$\DeclareMathOperator\perm{perm}$Let $G$ be a finite group. Define the determinant $\det(G)$ of $G$ as the determinant of the character table of $G$ over $\mathbb{C}$ and define the permanent $\perm(G)$ of $G$ as the permanent of the character table of $G$ over $\mathbb{C}$. Note that due to the properties of the determinant and the permanent, this definition just depends on $G$ and not on the ordering of the conjugacy classes etc.

I'm not experienced with character theory but did some experiments with GAP on this and found nothing related in the literature, which motivates the following questions (sorry, in case they are trivial). Of course finite groups are dangerous and it is tested just for all finite groups of order at most $n \leq 30$ and some other cases, which might not be too good evidence for a question on finite groups.

Question 1: Are $\perm(G)$ and $\det(G)^2$ always integers?

I was able to prove this for cyclic groups. Since the character table of the direct product of groups is given by their Kroenecker product, one can conclude that $\det(G)^2$ is also an integer for all abelian groups $G$. Maybe there is a formula for the permanent of the Kroenecker product of matrices to conclduge that $\perm(G)$ is also an integer for all abelian groups or even better a more direct proof that question 1 is true at least for abelian groups. Note that $\det(G)$ is in general not an integer, even for cyclic groups.

Now call a finite group permfect in case $\perm(G)=0$.

Question 2: Is it true that all finite groups of order $n$ are permfect if and only if $n=4r+2$ for some $r \geq 2$?

Being permfect could be seen as having a high symmetry. It seems symmetric groups are permfect and for the alternating groups I only found $A_6$ to be permfect yet.

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    $\begingroup$ Pretty sure this can be cracked with the usual Galois-theoretical pincers: Invariance under field automorphisms should show that the numbers $\operatorname{per} G$ and $\left(\det G\right)^2$ are rationals, while the algebraic integrality of the character values shows that these numbers are algebraic integers. Hence, they are integers. $\endgroup$ – darij grinberg Jun 21 at 8:45
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    $\begingroup$ The "if" direction of your Question 2 might have to do with the fact that if $n = 4r+2$, then any group $G$ of order $n$ has a normal subgroup of order $2$ (by math.stackexchange.com/questions/225987/… ). This gives the character table some sort of irregular chessboard structure. Do we know something about how many conjugacy classes of $G$ belong to said subgroup? $\endgroup$ – darij grinberg Jun 22 at 9:03
  • $\begingroup$ @darijgrinberg Thanks, you probably mean index 2 instead of order 2. There are groups having normal subgroups of index 2 without permanent 0, for example the Klein four group. I do some more experiments with GAP on that. $\endgroup$ – Mare Jun 22 at 11:41
  • $\begingroup$ Oops, yes, index $2$ is what I meant. I suspect something more is needed. $\endgroup$ – darij grinberg Jun 22 at 11:44
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    $\begingroup$ Permfect is a great term. $\endgroup$ – Brian Hopkins Jun 25 at 13:03
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I won't discuss the permanent part of the question, but I think the other part can be done easily, even without Galois theory. Let $X = X(G)$ denote the character table of $G$ (rows indexed by complex irreducible characters of $G$, columns indexed by conjugacy classes of $G$). Then, by the orthogonality relations, $X\overline{X}^{T}$ is an integer diagonal matrix with $i$-th main diagonal entry $|C_{G}(x_{i})|$, where $x_{i}$ is a representative of the $i$-th conjugacy class.

Hence we have $|{\rm det}(X)|^{2} = \prod_{i}|C_{G}(x_{i})|.$ On the other hand, note that replacing $X$ by $\overline{X}$ gives a matrix with the same rows as $X$, but permuted (under some permutation of order one or two), since the complex conjugate of an irreducible character is an irreducible character. Hence $\overline{X} = PX$, where $P$ is a permutation matrix associated to a permutation $\sigma$ of order at most two. Thus $\overline{\rm{det}(X)} = {\rm sign}(\sigma) {\rm det}(X)$.

If $\sigma$ is an odd permutation, we then see that ${\rm det}(X)^{2}$ is a negative integer, while if $\sigma$ is an even permutation, we see that ${\rm det}(X)^{2}$ is a positive integer.

Hence we have the curious fact that ${\rm det}(X)^{2}$ is a positive integer if the number of complex irreducible characters of $G$ with Frobenius-Schur indicator zero is divisible by $4$, and is a negative integer if the number of complex irreducible characters of $G$ with Frobenius-Schur indicator zero is congruent to $2$ (mod $4$). The number of complex irreducible characters of $G$ with Frobenius-Schur indicator zero is just the number of complex irreducible characters which are not real-valued, but I write it as above to illustrate the link between the F-S indicator and the congruence (mod 4).

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    $\begingroup$ See also the MO question mathoverflow.net/q/78050. $\endgroup$ – Denis Serre Jun 27 at 19:56
  • $\begingroup$ @DenisSerre : Thanks for pointing that out. I see that your first proof is essentially the same as that above. $\endgroup$ – Geoff Robinson Jun 27 at 20:20
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I'll address the remaining question 2.

Theorem: If $|G|=4r+2$ then the permanent of its character table vanishes.

Proof: We know $G$ has a normal subgroup $H$ of index $2$. Let's denote the cosets of $G/H$ by $\{H, aH\}$. Start by noticing that the conjugacy classes of $G$ either completely lie inside $H$ or inside $aH$. Let $k$ be the number of conjugacy classes that lie inside $aH$.

Next, we let $U$ be the 1 dimensional representation where elements of $H$ act as $1$ and elements of $aH$ act as $-1$. Tensoring with $U$ gives an involution on irreducible representations of $G$ and thus also an involution $\sigma$ of the columns of the character table.

Denoting by $S$ the set of all irreducible representations of $G$, and by $f$ the possible bijections between $S$ and the conjugacy classes we see that our permanent can be written as $$\sum_f \prod_{V\in S}\chi_V(f(V))=\sum_f\prod_{V\in S} \chi_{\sigma(V)} (f(V))=(-1)^k\sum_f \prod_{V\in S}\chi_V(f(V))$$ so the theorem will follow once we establish that $k$ is odd.

There is probably a better way of doing this but I reasoned as follows: $k$ is equal to the number of conjugacy classes that lie inside $H$ that don't split when we restrict to the conjugation action by $H$ (see lemma 2 here for a more general statement). Now, if $x\in H$ and $C_G(x)$ denotes its centralizer, we have $[C_G(x):C_H(x)]\in\{1,2\}$.

If this index is $1$ then then the conjugacy class of $x$ splits into two conjugacy classes of equal size when we restrict to the conjugation by $H$. In particular the size of the conjugacy class of $x$ is even. If the index is $2$ then the conjugacy class doesn't split and its size is $(4r+2)/|C_G(x)|$ which is odd.

Since the total number of elements in $H$ is $2r+1$ there must be an odd number of nonsplit conjugacy classes, so $k$ is odd and we are done.

One also has to check that for other orders it is possible to find groups with nonvanishing permanent. For odd order one can take the cyclic group $\mathbb Z/n\mathbb Z$, and for order divisible by 4, I believe you can take a product of the Klein four group with a cyclic group.


I also wanted to mention that the permanent of the symmetric group is not always zero and it's first few values are recorded in OEIS, in particular this permanent for $S_8$ is $-20834715303936$. There you will find a reference by Schmidt and Simion that uses an argument similar to the one above for the case of the symmetric group/alternating subgroup to prove that the permanent of the symmetric group vanishes whenever half the number of non-self-conjugate partitions of $n$ is odd. In a follow-up note they show that this occurs infinitely often.

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  • $\begingroup$ I think I found arxiv.org/abs/1802.08001 showing that the permanent of an elementary abelian 2-group of order $2^n \geq 4$ is indeed nonzero (and divisible by $2^n-1$). Together with your answer, this should completely answer question 2. $\endgroup$ – Mare yesterday

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