# On permanents and determinants of finite groups

$$\DeclareMathOperator\perm{perm}$$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.

• 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. Jun 21 '20 at 8:45
• 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? Jun 22 '20 at 9:03
• @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.
– Mare
Jun 22 '20 at 11:41
• Oops, yes, index $2$ is what I meant. I suspect something more is needed. Jun 22 '20 at 11:44
• Permfect is a great term. Jun 25 '20 at 13:03

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 $$|\det(X)|^{2} = \prod_{i}|C_{G}(x_{i})|.$$ 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 $$\overline{\det(X)} = \operatorname{sign}(\sigma) \det(X)$$.

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

Hence we have the curious fact that $$\det(X)^{2}$$ 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).

• See also the MO question mathoverflow.net/q/78050. Jun 27 '20 at 19:56
• @DenisSerre : Thanks for pointing that out. I see that your first proof is essentially the same as that above. Jun 27 '20 at 20:20

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 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 the product of an elementary abelian 2-group with a cyclic group of odd order.

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.

• 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.
– Mare
Jul 1 '20 at 15:55
• A more group theoretic argument that $aH$ is the union of an odd number of classis is as follows. Notice that $H$ has odd order by assumption, and that we might as well assume that $a$ has order $2$. Then since $\langle a \rangle$ is a Sylow $2$-subgroup of $G$, we see that every element of $G \backslash H$ is $G$-conjugate to an element of the form $ab$ where $b$ has odd order and $ab = ba$. Furthermore, if $b,c \in C_{H}(a)$ have odd order, then $ab$ and $ac$ are $G$-conjugate if and only $b$ and $c$ are already conjugate in $C_{H}(a)$.... Jul 3 '20 at 11:19
• So the number of $G$ classes in $aH$ is the number of conjugacy classes of $C_{H}(a)$. This is odd because $C_{H}(a)$ has odd order and only the identity is conjugate to its inverse in a group of odd order. Jul 3 '20 at 11:22
• @GeoffRobinson Very nice argument! Jul 4 '20 at 18:06

For the sake of completeness, here is the answer to Question 1, part of which is missing from the other answers:

Proposition 1. Let $$G$$ be a finite group. Consider the representations of $$G$$ over $$\mathbb{C}$$. Let $$\det G$$ denote the determinant of the character table of $$G$$. (Note that this is only defined up to sign, since the order of the rows and of the columns of the character table can be chosen arbitrarily.) Let $$\operatorname*{perm}G$$ denote the permanent of the character table of $$G$$. Then, $$\left(\det G\right)^2$$ and $$\operatorname*{perm}G$$ are integers.

To prove this, we need the following lemmas:

Lemma 2. Let $$G$$ be a finite group. Then, there is a finite Galois field extension $$\mathbb{F}$$ of $$\mathbb{Q}$$ such that all irreducible representations of $$G$$ are defined over $$\mathbb{F}$$.

Proof of Lemma 2. There is a finite Galois field extension $$\mathbb{K}$$ of $$\mathbb{Q}$$ such that all irreducible representations of $$G$$ are defined over $$\mathbb{K}$$. Indeed, this is known as a splitting field of $$G$$; its existence is part of Theorem 9.2.6 in Peter Webb, A Course in Finite Group Representation Theory, 2016.

Consider this field extension $$\mathbb{K}$$. Let $$\mathbb{F}$$ be the Galois closure of $$\mathbb{K}$$ over $$\mathbb{Q}$$ (or any other finite field extension of $$\mathbb{Q}$$ that is Galois over $$\mathbb{Q}$$ and contains $$\mathbb{K}$$ as a subfield). Then, all irreducible representations of $$G$$ are defined over $$\mathbb{F}$$ (since they are defined over $$\mathbb{K}$$, but $$\mathbb{F}$$ contains $$\mathbb{K}$$ as a subfield). This proves Lemma 2. $$\blacksquare$$

Lemma 3. Let $$G$$ be a finite group. Let $$\mathbb{F}$$ be a field extension of $$\mathbb{Q}$$ such that all irreducible representations of $$G$$ are defined over $$\mathbb{F}$$. Let $$\chi:G\rightarrow\mathbb{F}$$ be an irreducible character of $$G$$. Let $$\gamma:\mathbb{F}\rightarrow\mathbb{F}$$ be a $$\mathbb{Q}$$-algebra automorphism of $$\mathbb{F}$$. Then, $$\gamma\circ \chi:G\rightarrow\mathbb{F}$$ is an irreducible character of $$G$$.

Proof of Lemma 3. This is a totally straightforward "isomorphisms preserve all relative properties of objects they are applied to" argument, but for the sake of completeness, let me spell it out (at least to some level of detail):

The map $$\chi$$ is an irreducible character of $$G$$, and thus is the character of an irreducible representation $$\rho$$ of $$G$$. Consider this $$\rho$$, and WLOG assume that $$\rho$$ is a representation over $$\mathbb{F}$$. (This can be assumed since all irreducible representations of $$G$$ are defined over $$\mathbb{F}$$.) Thus, $$\rho$$ is a group homomorphism from $$G$$ to $$\operatorname*{GL} \nolimits_{n}\left( \mathbb{F}\right)$$ for some $$n\geq1$$. Consider this $$n$$.

The $$\mathbb{Q}$$-algebra automorphism $$\gamma:\mathbb{F}\rightarrow\mathbb{F}$$ induces a group automorphism $$\widetilde{\gamma}:\operatorname*{GL} \nolimits_{n}\left( \mathbb{F}\right) \rightarrow\operatorname*{GL} \nolimits_{n}\left( \mathbb{F}\right)$$ that transforms each matrix in $$\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right)$$ by applying $$\gamma$$ to each entry of the matrix. The composition $$\widetilde{\gamma} \circ\rho:G\rightarrow\operatorname*{GL}\nolimits_{n}\left( \mathbb{F} \right)$$ is a group homomorphism (since $$\widetilde{\gamma}$$ and $$\rho$$ are group homomorphisms), and thus is a representation of $$G$$. Moreover, the character of this representation $$\widetilde{\gamma}\circ\rho$$ is $$\gamma \circ\chi$$ (since $$\operatorname*{Tr}\left( \widetilde{\gamma}\left( A\right) \right) =\gamma\left( \operatorname*{Tr}A\right)$$ for any matrix $$A\in\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right)$$). We shall now show that this representation $$\widetilde{\gamma}\circ\rho$$ is irreducible.

Indeed, let $$U$$ be a subrepresentation of $$\widetilde{\gamma}\circ\rho$$ -- that is, an $$\mathbb{F}$$-vector subspace of $$\mathbb{F}^{n}$$ that is invariant under the action of $$\widetilde{\gamma}\circ\rho$$. Consider the $$\mathbb{Q}$$-module isomorphism $$\overline{\gamma}:\mathbb{F}^{n}\rightarrow \mathbb{F}^{n}$$ that applies $$\gamma$$ to each coordinate of the vector. Since $$\gamma$$ is a $$\mathbb{Q}$$-algebra homomorphism, we can easily see that $$\left( \widetilde{\gamma}\left( A\right) \right) \left( \overline {\gamma}\left( v\right) \right) =\overline{\gamma}\left( Av\right)$$ for each $$A\in\operatorname*{GL}\nolimits_{n}\left( \mathbb{F}\right)$$ and each $$v\in\mathbb{F}^{n}$$. Thus, we can easily see that $$\overline{\gamma} ^{-1}\left( U\right)$$ is an $$\mathbb{F}$$-vector subspace of $$\mathbb{F} ^{n}$$ that is invariant under the action of $$\rho$$ (since $$U$$ is an $$\mathbb{F}$$-vector subspace of $$\mathbb{F}^{n}$$ that is invariant under the action of $$\widetilde{\gamma}\circ\rho$$). In other words, $$\overline{\gamma }^{-1}\left( U\right)$$ is a subrepresentation of $$\rho$$. Since $$\rho$$ is irreducible, this entails that either $$\overline{\gamma}^{-1}\left( U\right) =0$$ or $$\overline{\gamma}^{-1}\left( U\right) =\mathbb{F}^{n}$$. Since $$\overline{\gamma}$$ is an isomorphism, we thus conclude that either $$U=0$$ or $$U=\mathbb{F}^{n}$$.

Forget that we fixed $$U$$. We thus have shown that if $$U$$ is a subrepresentation of $$\widetilde{\gamma}\circ\rho$$, then either $$U=0$$ or $$U=\mathbb{F}^{n}$$. In other words, the representation $$\widetilde{\gamma }\circ\rho$$ is irreducible (since its dimension is $$n\geq1$$). Thus, its character is an irreducible character of $$G$$. In other words, $$\gamma\circ \chi$$ is an irreducible character of $$G$$ (since $$\gamma\circ\chi$$ is the character of $$\widetilde{\gamma}\circ\rho$$). This proves Lemma 3. $$\blacksquare$$

Proof of Proposition 1. Lemma 2 shows that there is a finite Galois field extension $$\mathbb{F}$$ of $$\mathbb{Q}$$ such that all irreducible representations of $$G$$ are defined over $$\mathbb{F}$$. Consider this $$\mathbb{F}$$. Let $$\Gamma$$ be the Galois group $$\operatorname*{Gal}\left( \mathbb{F}/\mathbb{Q}\right)$$ (which consists of all $$\mathbb{Q}$$-algebra automorphisms of $$\mathbb{F}$$). The Fundamental Theorem of Galois Theory shows that the invariant ring $$\mathbb{F}^{\Gamma}$$ is $$\mathbb{Q}$$.

Let $$\chi_{1},\chi_{2},\ldots,\chi_{r}$$ be all irreducible characters of $$G$$ (listed without repetition). Note that these characters are maps from $$G$$ to $$\mathbb{F}$$ (since all irreducible representations of $$G$$ are defined over $$\mathbb{F}$$).

Let $$c_{1},c_{2},\ldots,c_{r}$$ be the conjugacy classes of $$G$$ (listed without repetition).

Let $$\operatorname*{per}A$$ denote the permanent of any square matrix $$A$$.

Let $$C$$ be the matrix $$\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}\in\mathbb{F}^{r\times r}$$. This matrix $$C$$ is the character table of $$G$$ (for the ordering of characters given by $$\chi_{1} ,\chi_{2},\ldots,\chi_{r}$$ and the ordering of conjugacy classes given by $$c_{1},c_{2},\ldots,c_{r}$$). Thus, the definition of $$\operatorname*{perm}G$$ shows that $$\operatorname*{perm}G$$ is the permanent of $$C$$. In other words, $$\operatorname*{perm}G=\operatorname*{per}C$$.

Let $$\gamma\in\Gamma$$. Thus, $$\gamma$$ is a $$\mathbb{Q}$$-algebra automorphism of $$\mathbb{F}$$ (since $$\gamma\in\Gamma=\operatorname*{Gal}\left( \mathbb{F}/\mathbb{Q}\right)$$). We shall show that $$\gamma\left( \operatorname*{perm}G\right) =\operatorname*{perm}G$$.

The $$\mathbb{Q}$$-algebra automorphism $$\gamma:\mathbb{F}\rightarrow\mathbb{F}$$ induces a $$\mathbb{Q}$$-algebra automorphism $$\widetilde{\gamma}:\mathbb{F}^{r\times r}\rightarrow\mathbb{F}^{r\times r}$$ that transforms each matrix in $$\mathbb{F}^{r\times r}$$ by applying $$\gamma$$ to each entry of the matrix.

We define a map $$f:\left\{ 1,2,\ldots,r\right\} \rightarrow\left\{ 1,2,\ldots,r\right\}$$ as follows:

Let $$i\in\left\{ 1,2,\ldots,r\right\}$$. Then, $$\chi_{i}:G\rightarrow \mathbb{F}$$ is an irreducible character of $$G$$. Thus, Lemma 3 (applied to $$\chi=\chi_{i}$$) shows that $$\gamma\circ\chi_{i}:G\rightarrow\mathbb{F}$$ is an irreducible character of $$G$$. Hence, $$\gamma\circ\chi_{i}=\chi_{j}$$ for some $$j\in\left\{ 1,2,\ldots,r\right\}$$ (since $$\chi_{1},\chi_{2},\ldots ,\chi_{r}$$ are all irreducible characters of $$G$$). This $$j$$ is uniquely defined. We define $$f\left( i\right)$$ to be $$j$$.

Thus, we have defined a map $$f:\left\{ 1,2,\ldots,r\right\} \rightarrow \left\{ 1,2,\ldots,r\right\}$$ with the property that $$$$\gamma\circ\chi_{i}=\chi_{f\left( i\right) }\qquad\text{for each } i\in\left\{ 1,2,\ldots,r\right\} . \label{eq.darij1.1} \tag{1}$$$$

If two distinct elements $$i$$ and $$j$$ of $$\left\{ 1,2,\ldots,r\right\}$$ would satisfy $$f\left( i\right) =f\left( j\right)$$, then they would satisfy $$\gamma\circ\chi_{i}=\gamma\circ\chi_{j}$$ (by \eqref{eq.darij1.1} ) and therefore $$\chi_{i}=\chi_{j}$$ (since $$\gamma$$ is invertible), which would contradict the fact that $$\chi_{1},\chi_{2},\ldots,\chi_{r}$$ are distinct. Thus, two distinct elements $$i$$ and $$j$$ of $$\left\{ 1,2,\ldots ,r\right\}$$ always satisfy $$f\left( i\right) \neq f\left( j\right)$$. In other words, the map $$f$$ is injective. Hence, $$f$$ is a permutation (since $$f$$ is an injective map from $$\left\{ 1,2,\ldots,r\right\}$$ to $$\left\{ 1,2,\ldots,r\right\}$$).

Now, from $$C=\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}$$, we obtain \begin{align*} \widetilde{\gamma}\left( C\right) =\left( \gamma\left( \chi_{i}\left( c_{j}\right) \right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}=\left( \chi_{f\left( i\right) }\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}, \end{align*} since each $$i,j\in\left\{ 1,2,\ldots,r\right\}$$ satisfy \begin{align*} \gamma\left( \chi_{i}\left( c_{j}\right) \right) =\left( \gamma\circ \chi_{i}\right) \left( c_{j}\right) =\chi_{f\left( i\right) }\left( c_{j}\right) \qquad\left( \text{by \eqref{eq.darij1.1}}\right) . \end{align*} Thus, the matrix $$\widetilde{\gamma}\left( C\right)$$ is obtained from the matrix $$\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}$$ by permuting the rows (since $$f:\left\{ 1,2,\ldots,r\right\} \rightarrow\left\{ 1,2,\ldots,r\right\}$$ is a permutation). In other words, the matrix $$\widetilde{\gamma}\left( C\right)$$ is obtained from the matrix $$C$$ by permuting the rows (since $$C=\left( \chi_{i}\left( c_{j}\right) \right) _{1\leq i\leq r,\ 1\leq j\leq r}$$). Hence, $$\operatorname*{per} \left( \widetilde{\gamma}\left( C\right) \right) =\operatorname*{per}C$$ (since the permanent of a matrix does not change when its rows are permuted).

But the definition of $$\widetilde{\gamma}$$ yields that $$\operatorname*{per} \left( \widetilde{\gamma}\left( C\right) \right) =\gamma\left( \operatorname*{per}C\right)$$ (since $$\gamma$$ is a $$\mathbb{Q}$$-algebra homomorphism). Hence, $$\gamma\left( \operatorname*{per}C\right) =\operatorname*{per}\left( \widetilde{\gamma}\left( C\right) \right) =\operatorname*{per}C$$. In view of $$\operatorname*{perm}G=\operatorname*{per} C$$, this rewrites as $$\gamma\left( \operatorname*{perm}G\right) =\operatorname*{perm}G$$.

Forget that we fixed $$\gamma$$. We thus have shown that $$\gamma\left( \operatorname*{perm}G\right) =\operatorname*{perm}G$$ for each $$\gamma \in\Gamma$$. In other words, $$\operatorname*{perm}G$$ belongs to the invariant ring $$\mathbb{F}^{\Gamma}$$. In other words, $$\operatorname*{perm}G$$ belongs to $$\mathbb{Q}$$ (since the invariant ring $$\mathbb{F}^{\Gamma}$$ is $$\mathbb{Q}$$).

But all values of the characters $$\chi_{1},\chi_{2},\ldots,\chi_{r}$$ are sums of roots of unity (because they are traces of matrices $$A\in\operatorname*{GL} \nolimits_{n}\left( \mathbb{F}\right)$$ that satisfy $$A^{\left\vert G\right\vert }=I_{n}$$, and the eigenvalues of such a matrix are roots of unity), and thus are algebraic integers. Hence, all entries of the matrix $$C$$ are algebraic integers (since all these entries are values of the characters $$\chi_{1},\chi_{2},\ldots,\chi_{r}$$). Thus, the permanent $$\operatorname*{per} C$$ of this matrix $$C$$ is an algebraic integer (since the algebraic integers form a ring). In other words, $$\operatorname*{perm}G$$ is an algebraic integer (since $$\operatorname*{perm}G=\operatorname*{per}C$$). Hence, $$\operatorname*{perm}G$$ is an algebraic integer in $$\mathbb{Q}$$ (since $$\operatorname*{perm}G$$ belongs to $$\mathbb{Q}$$). Since the only algebraic integers in $$\mathbb{Q}$$ are integers (because the ring $$\mathbb{Q}$$ is integrally closed), this entails that $$\operatorname*{perm}G$$ is an integer.

A similar argument shows that $$\left( \det G\right) ^{2}$$ is an integer. (Here we need to use the fact that the square of the determinant of a matrix does not change when its rows are permuted. This is because the determinant gets multiplied by a power of $$-1$$.) Thus, Proposition 1 is proved. $$\blacksquare$$

• Note that use of the transpose inverse (dual) representation shows that the complex conjugate of an irreducible character is an irreducible character. Also, as a matter of interest, it is not necessary to have the realizability theorem of the representations over a suitable cyclotomic field to know that algebraic conjugates of irreducible characters are irreducible characters. The irreducible characters of $G$ are (by Schur's Lemma) in bijection with the one-dimensional representations of $Z(\mathbb{C}G)$, the center of the complex group algebra. Aug 31 '20 at 13:49
• Since the conjugacy class sums form a basis for $Z(\mathbb{C}G)$ and products of class sums are integer combinations of class sums, it follows that the linear characters of $Z(\mathbb{C}G)$ take algebraic integer values on class sums, and that algebraic conjugates of linear characters of $Z(\mathbb{C}G)$ are still linear character of $Z(\mathbb{C}G)$. Aug 31 '20 at 13:55
• @GeoffRobinson: "The irreducible characters of $G$ are (by Schur's Lemma) in bijection with the one-dimensional representations of $Z\left(\mathbb{C}G\right)$, the center of the complex group algebra." Nice observation! (Although I don't see it as a consequence of Schur's lemma, but rather as a consequence of the Artin-Wedderburn theorem.) I'm not sure to what extent it could shorten my proof, though -- I still need a splitting field, don't I? (Composing a group representation with a Galois automorphism of the field clearly yields a group representation, but composing ... Aug 31 '20 at 14:05
• In general, composing a linear representation of a commutative with a Galois automorphism does not lead to another representation, but, as I indicated, the special basis of class sums for the center of the group algebra (which multiply to integral combinations of class sums) means that an algebraic conjugate of a linear character leads naturally to another linear character, in the following. The (non-zero) linear function $\lambda: Z(\mathbb{C}G) \to \mathbb{C}$ yields an algebra homomorphism if and only if we have $\lambda(C_{i}) \lambda(C_{j}) = \sum_{m=1}^{k} a_{ijm} \lambda(C_{m})$ Aug 31 '20 at 17:09
• .whenever we have $C_{i}C_{j} = \sum_{m=1}^{k}a_{ijm}C_{m}$, where $C_{i},C_{j},C_{m}$ are class sums, and the $a_{ijm}$ are the (rational integer) class algebra constants. The $\lambda(C_{i})$ are always algebraic integers (even cyclotomic integers), and it is clear that if $\lambda$ satisfies this property, so does $\lambda^{\sigma}$ (defined on class sums), when $\sigma$ is a Galois automorphism. Burnside and Frobenius knew this fact I think. Aug 31 '20 at 17:15