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Call a number $n$ special in case there is a unique group of order n whose character table consists of only integers. This should be equivalent to the condition that the number of conjugacy classes coincides with the number of conjugacy classes of cyclic subgroups.

Questions: What is the sequence of special numbers? It starts with 1,2, 4, 6, 12, 18, 54,120. Is it an infinite sequence?

For which $m$ is $m!$ special ?(note that the symmetric group has integral character table and thus $n!$ is a good candidate for being special)

[Side question: For which m are the groups of order m! classified?]

What are the corresponding unique groups of order $n$ for $n$ being special?

The sequence of such groups starts with $\mathbb{Z}_1, \mathbb{Z}_2, \mathbb{Z}_2 \times \mathbb{Z}_2, S_3,D_{12}, (\mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2,(\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3) \rtimes \mathbb{Z}_2, S_5$

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    $\begingroup$ I don't get the definition. What is $n$? Is it the order of the group? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Nov 4, 2018 at 13:15
  • $\begingroup$ @JoséFigueroa-O'Farrill Yes. Thanks I forgot that and added it now. $\endgroup$
    – Mare
    Commented Nov 4, 2018 at 13:16
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    $\begingroup$ I believe that Pal Hegedus proved that solvable groups with a rational character table are all $\{2,3,5\}$-groups, with an elementary Abelian normal Sylow $5$-subgroup. It is not clear to me at the moment which numbers $n$ of the form $n= 2^{a}3^{b}5^{c}$ have a chance of being special. $\endgroup$
    – Geoff Robinson
    Commented Nov 4, 2018 at 15:31

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