# Is a finite group given by its character table if its Sylow subgroups are so?

As pointed out by Mikko Korhonen in this answer, Özdem Çelik proved (in 1976 here) that a finite group whose Sylow subgroups are cyclic (called a Z-group) is determined by its character table.

Now there are many results and conjectures relating character tables and Sylow subgroups (see this paper of Gabriel Navarro), the most famous being perhaps the McKay conjecture.

This leads to wonder whether Çelik's theorem can be extended*.

Question 1: Is a finite group determined by its character table only if its Sylow subgroups are so?

Question 2: Is a finite group not in a Brauer pair only if its Sylow subgroups are so?
(negative answer suspected by Alex B.)

*Question 3: Is a finite group determined by its character table if its Sylow subgroups are so?
(it is this question which wonders whether Çelik's theorem can be extended)

Question 4: Is a finite group not in a Brauer pair if its Sylow subgroups are so?

• Çelik's paper MR0470050 is old and very hard to find. Can anyone having a copy put it in comment? – Sebastien Palcoux Mar 30 '19 at 12:54

The answer to the first question is negative. The group $${\rm SL}_2(\mathbb{F}_3)$$ has a $$2$$-Sylow subgroup isomorphic to $$Q_8$$, which is not determined by its character table, but $${\rm SL}_2(\mathbb{F}_3)$$ is the only group with its character table.
One way of seeing how the counterexample above worked is this: the group $$Q_8$$ has an outer automoprhism of order $$3$$ (cyclicly permuting $$i$$, $$j$$, and $$k$$), and $${\rm SL}_2(\mathbb{F}_3)$$ is a semidirect product of $$Q_8$$ and $$C_3$$, with the action given by this automorphism. On the other hand, $$D_8$$ has no automorphisms of order $$3$$, so one cannot produce an analogous construction with this cousin of $$Q_8$$. That makes it "unlikely" (in no precise sense) that there is another group with the same character table as $${\rm SL}_2(\mathbb{F}_3)$$.
I have checked that the group that Lux and Pahlings call $$G_{3378}$$ in Theorem 2.6.2 has an automorphism $$\alpha$$ of order $$3$$, while the other group in the Brauer pair, $$G_{3380}$$, does not. So I would conjecture that the group $$G_{3378}\rtimes\langle \alpha\rangle$$ is determined by its character table. This is a group of order $$768$$, so magma has the complete list of all groups of this order, and the conjecture should not be too hard to check, but I have not done that.
• We can deduce the Grothendieck ring of a finite group from its character table. Now a finite group $G$ is simple iff its Grothendieck ring is simple. So, if there is no simple group of order $|G|$ non-isomorphic to $G$ then $G$ is determined by its character table (in fact it's true for any finite simple group by this answer). So any finite simple group with a Sylow not determined by its character table will be a counter-example to Q1: $PSL_2(7) \supset D_8$, $A_6 \supset D_8$. For Q2 we need to find a simple group with a Sylow in a Brauer pair. – Sebastien Palcoux Mar 30 '19 at 19:25