# Finite groups with the same character table *including* class types, and square-free order

There are non-isomorphic finite groups with the same (complex) character table, as $$D_4$$ and $$Q_8$$.

$$\scriptsize\begin{array}{c|c} \text{class}&1&2A&2B&2C&4 \newline \text{size}&1&1&2&2&2 \newline \hline \rho_1 &1&1&1&1&1 \newline \rho_2 &1&1&-1&1&-1 \newline \rho_3 &1&1&1&-1&-1 \newline \rho_4 &1&1&-1&-1&1 \newline \rho_5 &2&-2&0&0&0 \newline \end{array} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{array}{c|c} \text{class}&1&2&4A&4B&4C \newline \text{size}&1&1&2&2&2 \newline \hline \rho_1 &1&1&1&1&1 \newline \rho_2 &1&1&-1&1&-1 \newline \rho_3 &1&1&1&-1&-1 \newline \rho_4 &1&1&-1&-1&1 \newline \rho_5 &2&-2&0&0&0 \newline \end{array}$$ But the character tables of $$D_4$$ and $$Q_8$$ are no more equal if we include the class types, as $$(1,2A,2B,2C,4) \neq (1,2,4A,4B,4)$$. A class is of type $$nX$$ if its elements has order $$n$$.

Question 1: Are there non-isomorphic groups with the same character table including class types?

Answer: Yes (see the comment of Derek Holt) and if in addition their conjugacy classes have the same power map, they are called Brauer pairs. Among the $$2$$-groups, the smallest order of a group in a Brauer pair is $$2^8$$, and among the $$56092$$ groups of order $$2^8$$, there are exactly ten Brauer pairs (see MR2680716 Theorem 2.6.2 page 136).

I am specifically interested in groups of square-free order.

Question 2: Is there a Brauer pair of square-free order groups?
(unless any two square-free order groups with same character table are isomorphic)

• Yes, if you search for Brauer pairs you will find plenty of references. – Derek Holt Mar 27 '19 at 11:25
• @DerekHolt: Thanks! The examples I found are all $p$-groups. I need to know if there are examples of square-free order. – Sebastien Palcoux Mar 27 '19 at 12:02
• @MikkoKorhonen perhaps you should make your comment into an answer. – Derek Holt Mar 30 '19 at 5:36

A finite group with all Sylow subgroups cyclic is called a $$Z$$-group.
According to review MR0470050 in MathSciNet, in [1] it is shown that a $$Z$$-group is determined by its character table. So the answer to question 2 would be no.
However, perhaps with some effort you could figure out your own proof. We have a good understanding of the structure of $$Z$$-groups (in particular of groups of squarefree order). See for example [2] for an enumeration of all $$Z$$-groups of given order. I would guess the character tables of $$Z$$-groups are also known.