Let $G$ be a finite group of order $n$ and $kG$ the group ring over a field of characteristic $0$. Let $C$ denote the Cayley table. The determinant of $C$ in $kG$ is defined as $det(C)=\sum_{\sigma \in S_n}sgn(\sigma)\, C_{1,\sigma(1)}\cdot ... \cdot C_{n,\sigma(n)}$. Does this always vanish? Clearly, $C$ is equivalent to the 'skew-symmetric' matrix $\tilde{C}_{g,h}=gh^{-1}$, but still I have no good argument.. but maybe it's wrong, I checked it only until $n=4$..


2 Answers 2


You address a question that has been historically very important. You will find answers there.

The first reference given there is a very nice book from American Mathematical Society that I recommend.

Another simpler reference is to be found in "Mathematical Conversations: Selections from The Mathematical Intelligencer" by Robin Wilson and Jeremy Gray, Springer 2012 pages 128-140 (partly available as a Google book).

As you, instead of taking directly the (Cayley) table of the group (i.e. taking entry $(k,l)=a_ka_l$, they take it to be $a_k(a_l)^{-1}.$ I borrow an example given in the last cited book, dealing with $S_3$, the symmetric group on 3 elements:

Let $g_1=1, \ g_2=(1 2), \ g_3=(2 3), \ g_4=(1 3), \ g_5=(1 2 3), \ g_6=(1 2 3).$

Let $x_i=x_{g_i}$. The group determinant of $S_3$ is:

$$det(x_{g_ig_j^{-1}})=\begin{vmatrix}x_1&x_2&x_3&x_4&x_6&x_5\\x_2&x_1&x_5&x_6&x_4&x_3\\x_3&x_6&x_1&x_5&x_2&x_4\\x_4&x_5&x_6&x_1&x_3&x_2\\x_5&x_4&x_2&x_3&x_1&x_6\\x_6&x_3&x_4&x_2&x_5&x_1\end{vmatrix}=F_1F_2(F_3)^2 \ \ \ \text{with}$$

$$\cases{F_1=x_1+x_2+x_3+x_4+x_5+x_6\\F_2=x_1-x_2-x_3-x_4+x_5+x_6\\ F_3=x_1^2-x_2^2+x_2x_3-x_3^2+x_2x_4+x_3x_4-x_4^2-x_1x_5+x_5^2-x_1x_6-x_5x_6+x_6^2}$$

These different factors give specific information;

  • for example, in $F_2$, the alternation of signs + - - - + + corresponds to "signature" (identity has signature 1, the signature of the three transpositions is -1, and the two cycles have signature 1).(see paragraph 5.2 in (www.math.ubc.ca/~carrell/Book2_Sn.pdf))

  • the exponents $1$ for $F_1$ and $F_2$, two for $F_3$ is linked to the dimensionality of the corresponding representation ($F_3$ is associated with a two dimensional matrix representation).

See around p. 420 a well written article in the London Mathematical Society Lecture Notes series 261 (Groups St Andrews 1997 II), Campbell et al, editors available as a Google Book. A similar material, by the same author Kenneth W. Johnson, can be found to his recent book completely devoted to "Group Matrices, Group Determinants and Representation Theory", viewable as well as a Google book.

As said in this book, these results have been extended to latin squares that are not necessarily group tables;

See also this [reference] (https://kconrad.math.uconn.edu/articles/groupdet.pdf)

also this

  • $\begingroup$ Does it matter whether the group is abelian or not? Just out of curiousity. $\endgroup$
    – snulty
    Commented Nov 18, 2016 at 21:31
  • $\begingroup$ I think that for an abelian group, the factors are all first degree. But I must check... $\endgroup$
    – Jean Marie
    Commented Nov 18, 2016 at 23:16
  • $\begingroup$ Thank you, this seems like a very cool theory! I'm not sure but does this imply an answer to my question (so what happens if you replace the polynomial ring by the group ring)? $\endgroup$ Commented Nov 19, 2016 at 0:02

You might want to read a fantastic overview of the theory of representIons of finite groups by prof T. Y. Lam: http://www.ams.org/notices/199803/lam.pdf Look for the group determinant.


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