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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)

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

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.

I do not have access to [1] and I have not seen it, if you are interested it seems you might have to look for a physical copy in a library somewhere.

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.

[1] Çelik, Özdem. On the character table of Z-groups. Mitt. Math. Sem. Giessen Heft 121 (1976), 75–77

[2] M.R. Murty and V.K. Murty. On groups of squarefree order. Math. Ann. 267 (1984), 299–309.

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