The character table of a finite group will be called integral if all its entries are integers. There are $11$ such groups up to order $16$, namely $C_1$, $C_2$, $C_2^2$, $S_3$, $D_8$, $Q_8$, $C_2^3$, $D_{12}$, $C_2 \times D_8$, $C_2 \times Q_8$ and $C_2^3$. There are $76$ such groups up to order $120$ (see Appendix), the last one of the list, namely $S_5$, is of specific interest as it is the first non-solvable one. In fact, every symmetric group has an integral character table. It can be proved using the following result (Lemma 7.15 in this note by Sam Raskin):

Lemma: Let $G$ be a finite group such that for all $g \in G$ of order $n$ and for all $m$ coprime to $n$, $g^m$ is conjugate to $g$. Then $G$ has an integral character table.

Question: What is known about the finite groups with integral character table? Are they classified?

I mainly ask for references reviewing what is known about such groups. Otherwise here is a list of questions:

  • Is the converse of above lemma true?
  • I guess that $C_2$ is the only such group which is simple. Is it true?
  • I guess that $C_1$ is the only such group of odd order. Is it true?
  • Is there such a group with a non-abelian and non-alternating simple normal subgroup?
  • Is every finite group a normal subgroup of such a group?

Appendix: List of all the finite groups with integral character table up to order $120$.

gap> for o in [1..120] do
>       if o=1 then Print("\n","|G| ","Nr ","G ","\n","\n");fi;
>       n:=NrSmallGroups(o);;
>       for i in [1..n] do
>               G:=SmallGroup(o,i);;
>               irr:=Irr(G);;
>               s:=Size(irr);;
>               c:=0;;
>               for j in [1..s] do
>                       if Conductor(irr[j])<>1 then
>                               c:=1;;
>                               break;
>                       fi;
>               od;
>               if c=0 then
>                       Print(o,"   ",i,"   ",StructureDescription(G),"\n");
>               fi;
>       od;
> od;

|G| Nr G

1   1   1
2   1   C2
4   2   C2 x C2
6   1   S3
8   3   D8
8   4   Q8
8   5   C2 x C2 x C2
12   4   D12
16   11   C2 x D8
16   12   C2 x Q8
16   14   C2 x C2 x C2 x C2
18   4   (C3 x C3) : C2
24   12   S4
24   14   C2 x C2 x S3
32   27   (C2 x C2 x C2 x C2) : C2
32   34   (C4 x C4) : C2
32   35   C4 : Q8
32   43   C8 : (C2 x C2)
32   44   (C2 x Q8) : C2
32   46   C2 x C2 x D8
32   47   C2 x C2 x Q8
32   49   (C2 x C2 x C2) : (C2 x C2)
32   50   (C2 x Q8) : C2
32   51   C2 x C2 x C2 x C2 x C2
36   10   S3 x S3
36   13   C2 x ((C3 x C3) : C2)
48   38   D8 x S3
48   40   Q8 x S3
48   48   C2 x S4
48   51   C2 x C2 x C2 x S3
54   14   (C3 x C3 x C3) : C2
64   134   ((C4 x C4) : C2) : C2
64   137   (C4 : Q8) : C2
64   138   ((C2 x C2 x C2 x C2) : C2) : C2
64   177   ((C4 x C4) : C2) : C2
64   178   (C4 : Q8) : C2
64   182   C8 : Q8
64   202   C2 x ((C2 x C2 x C2 x C2) : C2)
64   211   C2 x ((C4 x C4) : C2)
64   212   C2 x (C4 : Q8)
64   215   (C2 x C2 x D8) : C2
64   216   (C2 x ((C4 x C2) : C2)) : C2
64   217   ((C4 x C4) : C2) : C2
64   218   (C2 x ((C4 x C2) : C2)) : C2
64   224   ((C2 x Q8) : C2) : C2
64   225   (C4 : Q8) : C2
64   226   D8 x D8
64   230   Q8 x D8
64   239   Q8 x Q8
64   241   ((C4 x C2 x C2) : C2) : C2
64   242   ((C4 x C4) : C2) : C2
64   243   ((C4 x C2 x C2) : C2) : C2
64   244   (C4 : Q8) : C2
64   245   (C2 x C2) . (C2 x C2 x C2 x C2)
64   254   C2 x (C8 : (C2 x C2))
64   255   C2 x ((C2 x Q8) : C2)
64   261   C2 x C2 x C2 x D8
64   262   C2 x C2 x C2 x Q8
64   264   C2 x ((C2 x C2 x C2) : (C2 x C2))
64   265   C2 x ((C2 x Q8) : C2)
64   267   C2 x C2 x C2 x C2 x C2 x C2
72   40   (S3 x S3) : C2
72   41   (C3 x C3) : Q8
72   43   (C3 x A4) : C2
72   46   C2 x S3 x S3
72   49   C2 x C2 x ((C3 x C3) : C2)
96   209   C2 x D8 x S3
96   212   C2 x Q8 x S3
96   226   C2 x C2 x S4
96   227   ((C2 x C2 x C2 x C2) : C3) : C2
96   230   C2 x C2 x C2 x C2 x S3
108   17   ((C3 x C3) : C3) : (C2 x C2)
108   39   ((C3 x C3) : C2) x S3
108   44   C2 x ((C3 x C3 x C3) : C2)
120   34   S5
  • 2
    $\begingroup$ I think Weyl groups are known to be examples. Using the Weyl group of $E_8$ you get an example with a nonabelian simple normal subgroup that is not an alternating group. $\endgroup$
    – YCor
    Aug 17 at 8:20
  • 1
    $\begingroup$ Answer to your third question: yes $1$ is the only example of odd order. If $G$ has this property and has odd order $\ge 1$, let $C$ be a cyclic subgroup of prime order $p$. Then the normalizer of $C$ is transitive on $C-\{1\}$, so its order is divisible by $p-1$, hence is even. (This also proves that whenever a prime $p$ divides $|G|$, then $p-1$ divides $|G|$.) $\endgroup$
    – YCor
    Aug 17 at 8:23
  • $\begingroup$ You should be aware that the GAP structure description of a group does not always identify a group up to isomorphism. It would be more informative to print the numbers of the groups of each order. $\endgroup$
    – Derek Holt
    Aug 17 at 8:23
  • 1
    $\begingroup$ My first guess is false, see Corollary B.1 in On finite rational groups and related topics by Feit and Seitz: Let $G$ be a noncyclic simple group. Then $G$ has an integral character table iff $G =Sp_6(2)$ or $O_8^+(2)’$. $\endgroup$ Aug 17 at 9:32
  • 1
    $\begingroup$ oeis.org/A064527 Numbers k such that there exists a finite group G of order k such that all entries in its character table are integers. $\endgroup$ Aug 17 at 21:39

2 Answers 2


There is no complete classification, but some structural results are known. To give you something to search for: such groups are called $\mathbb{Q}$-groups. There is a whole book devoted to their structure: Structure and Representations of $\mathbb{Q}$-Groups by D. Kletzing. You will find there answers to many, if not all, of your questions.

In particular, the converse of the lemma is indeed true. It follows from two facts: firstly, characters "separate" conjugacy classes, i.e. if two elements are not conjugate, then there exists an irreducible character that takes different values on them; and secondly, if $m$ is coprime to the order of $g$, then for every irreducible character $\chi$ the values $\chi(g)$ and $\chi(g^m)$ are Galois conjugates.

  • 1
    $\begingroup$ I found a very strong result in the book your cited (Corollary 12): let $G$ be a $\mathbb{Q}$-group and let $g \in G$, then $ord(g)$ divides $|G|$. In particular, for every prime divisor $p$ of $|G|$ then $p-1$ divides $|G|$. So if $G$ is not trivial then it must be of even order. $\endgroup$ Aug 17 at 8:16
  • $\begingroup$ Oh, I've just mentioned this fact above (I didn't know this fact but it's a straightforward observation). $\endgroup$
    – YCor
    Aug 17 at 8:26
  • 2
    $\begingroup$ In my comment above, $ord(g)$ should be replace by $\varphi(ord(g))$, otherwise it is just Lagrange theorem... $\endgroup$ Aug 17 at 21:08

It was proved by W. Feit and G. Seitz that there are (at most) five non-Abelian simple groups which are not alternating and may occurs as composition factors of groups with integral character table (to use your terminology). There is a reference to their paper in the 2008 paper of J.G. Thompson "On composition factors of rational finite groups" ( Journal of Algebra, 319 (2008), 558-594.

In his paper, Thompson proves that if a cyclic group of order $p$ occurs as a composition factor of a rational finite group, then $p \leq 11$, and expresses the belief that the correct bound may be $p \leq 5$.

Work on solvable rational groups has been done by R. Gow, and later by P. Hegedus.

Later edit: A more general supply of rational finite groups than symmetric groups is provided by finite real reflection groups, which usually have all complex characters rational-valued.

  • 3
    $\begingroup$ The five non-abelian simple groups mentioned are $PSp_4(3)$, $Sp_6(2)$, $PSL_3(4)$, $PSU_4(3)$, and $\Omega_8^{+}(2)$. $\endgroup$
    – spin
    Aug 17 at 14:28

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