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
Best Answer
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.