Let $a\mathop{.}b \stackrel{\text{def}}{=} aba^{-1} $ denote conjugation by $a$

Suppose we define a matrix $M$, the "conjugation table", associated with our finite group $G = (X,*_{\small{G}})$ as follows. (I'm considering the cells of $M$ to be formal sums of group elements (with the product of monomials defined in terms of the group operation), but I'm only using that machinery to talk about equivalence up to relabeling.)

$$ M_{ij} \stackrel{\text{def}}{=} x_i \mathop{.} x_j = x_i x_j x_i^{-1} $$

I'm also thinking of two matrices $M$ and $M'$ as equivalent if they only differ by a permutation / relabelling, so

$$ M \sim M' \stackrel{\text{def}}{\iff} MP=M' \;\;\text{where $P$ is a permutation matrix} $$

or equivalently

$$ M \sim M' \stackrel{\text{def}}{\iff} M_{ij} = M'_{\sigma i \sigma j} \;\;\text{where $\sigma$ is a permutation} $$

I can think of a case where an $M$ does not uniquely identify a group and a case where an $M$ does uniquely identify a group.

I think a group is Abelian if and only if the following holds. (The "if" direction is trivial).

$$ x_i \mathop{.} x_j = x_j \;\;\forall i,j $$

So, if $G$ has four elements and is Abelian, then it could be the cyclic group on four elements $Z_4$ or the Klein four group $V_4$.

$Z_4$ and $V_4$ are indistinguishable by their "conjugation tables".

However, if $G$ has three elements, it can only be $Z_3$ since there's only one group of order 3.

So, there are at least some circumstances under which a given $M$ is associated with exactly one group. Do we know what those circumstances are?


Basically, what you call "conjugation table" is the left $G$-set structure on $X$ that is defined by conjugation. The equivalence is nothing but isomorphism as $G$-sets.

From the $G$-set structure on $X$, several invariants of $G$ can be calculated.

  • the order of $G$ = the size of $X$
  • the number of conjugacy classes of $G$ = the number of $G$-orbits in $X$
  • the sizes of conjugacy classes $\{\, \lvert g^G \rvert : g \in G \,\}$ = the sizes of $G$-orbits in $X$ (as multisets)
  • the sizes of centralizers $\{\, \lvert C_G(g) \rvert : g \in G \,\}$ = the sizes of stabilizers $\{\, \lvert G_x \rvert : x \in X \,\}$ (as multisets)
  • the inner automorphism group $\operatorname{Inn}(G)$ = the automorphism group $\operatorname{Aut}_G(X)$ as $G$-set

So one of the simplest circumstance "when a conjugation table determines a finite group" is when $G$ has trivial center $Z(G) = 1$. This is because $\operatorname{Inn}(G) \cong G/Z(G)$ in general.


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