# Determinant of Cayley table

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 $n=4$..

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;

also this

• Does it matter whether the group is abelian or not? Just out of curiousity. Nov 18, 2016 at 21:31
• I think that for an abelian group, the factors are all first degree. But I must check... Nov 18, 2016 at 23:16
• 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)? 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.