[GAP Forum] Print the irreducible characters of the normal ordinary character table format.

[Hongyi Zhao]

On Sat, Apr 30, 2022 at 7:15 PM Dima Pasechnik <dima at sagemath.org> wrote:
>
> On Sat, Apr 30, 2022 at 05:36:59PM +0800, Hongyi Zhao wrote:
> > On Sat, Apr 30, 2022 at 4:44 PM Bill Allombert
> > <Bill.Allombert at math.u-bordeaux.fr> wrote:
> > >
> > > On Sat, Apr 30, 2022 at 03:21:10PM +0800, Hongyi Zhao wrote:
> > > > Hi GAP team,
> > > >
> > > > I use the following code snippet to compute the irreducible characters
> > > > of a finitely presented group:
> > > >
> > > > So, I want to get a pretty printed result with the rows correspond to
> > > > irreducible representations, and the columns correspond to conjugacy
> > > > classes of group elements.
> > > >
> > > > Are there any clues to achieve this goal?
> > >
> > > Use Display(CharacterTable(g));
> > >
> > > gap> G:=CyclicGroup(8);
> > > <pc group of size 8 with 3 generators>
> > > gap> Display(CharacterTable(G));
> > > CT1
> > >
> > >      2  3   3  3  3   3   3  3   3
> > >
> > >        1a  8a 4a 2a  8b  8c 4b  8d
> > >
> > > X.1     1   1  1  1   1   1  1   1
> > > X.2     1  -1  1  1  -1  -1  1  -1
> > > X.3     1   A -1  1  -A   A -1  -A
> > > X.4     1  -A -1  1   A  -A -1   A
> > > X.5     1   B  A -1 -/B  -B -A  /B
> > > X.6     1  -B  A -1  /B   B -A -/B
> > > X.7     1 -/B -A -1   B  /B  A  -B
> > > X.8     1  /B -A -1  -B -/B  A   B
> > >
> > > A = E(4)
> > >   = Sqrt(-1) = i
> > > B = E(8)
> >
> > See the following results in my example
> >
> > gap> f := FreeGroup( "p", "q");;
> > gap> g42:= f/[ [ f.1 , f.1^-1 ], [ f.2 , f.2^-1 ], [ f.2 *f.1, f.1 *f.2 ] ];
> > <fp group on the generators [ p, q ]>
> >
> >
> > gap> Display(CharacterTable(g42));
> > CT1
> >
> >      2  2  2  2  2
> >
> >        1a 2a 2b 2c
> >     2P 1a 1a 1a 1a
> >
> > X.1     1  1  1  1
> > X.2     1 -1 -1  1
> > X.3     1 -1  1 -1
> > X.4     1  1 -1 -1

Very strange. Now, I run the same commands sequences as above, but
obtained the following different output:

gap> f := FreeGroup( "p", "q");;
gap> g42:= f/[ [ f.1 , f.1^-1 ], [ f.2 , f.2^-1 ], [ f.2 *f.1, f.1 *f.2 ] ];
<fp group on the generators [ p, q ]>
gap> Display(CharacterTable(g42));
CT1

     2  2  2  2  2

       1a 2a 2b 2c

X.1     1  1  1  1
X.2     1 -1 -1  1
X.3     1  1 -1 -1
X.4     1 -1  1 -1

As you can see, this time, the following line doesn't appear in the output:

2P 1a 1a 1a 1a


> >
> > Now, I'm confused on the following lines shown above:
> >
> >      2  2  2  2  2
> >
> >        1a 2a 2b 2c
> >     2P 1a 1a 1a 1a
> >
> >
> > 1. What's the meaning of all 2's in this line?
> >
> >      2  2  2  2  2
>
> the 1st 2 means "for prime divisor 2", and the other 2s indicate that
> these elements (conjugacy classes, more precisely) all have centralisers of order 2^2.

Sorry, my knowledge of group theory is still very rudimentary. So, I
try to describe my understanding as follows:

In this case, I can list the conjugacy classes as follows:

gap> ConjugacyClasses(g42);
[ <identity ...>^G, p^G, q^G, p*q^G ]

As for the statement given by you "... all have centralisers of order
2^2", how can I check it with GAP commands to validate and confirm, so
that I can have intuitive cognition and understanding?

> >
> > 2. What's the meaning of 1a, 2a, 2b, 2c, and 2P, respectively?
>
> 2P means "2nd Power". 1a, 2a, etc are names of conjugacy classes.

So, how do I know which group elements are included in the
corresponding conjugate class?

>
> Let's look at more generic character table, of the alternating group G=A_6 of order 360.

I confirmed as follows:

gap> Order(AlternatingGroup(6));
360

>
> gap> Display(CharacterTable("A6"));
>
>      2  3  3  .  .  2  .  .
>      3  2  .  2  2  .  .  .
>      5  1  .  .  .  .  1  1
>
>        1a 2a 3a 3b 4a 5a 5b
>     2P 1a 1a 3a 3b 2a 5b 5a
>     3P 1a 2a 1a 1a 4a 5b 5a
>     5P 1a 2a 3a 3b 4a 1a 1a
>
> X.1     1  1  1  1  1  1  1
> X.2     5  1  2 -1 -1  .  .
> [...]
>
> in the top portiion of the table you see the 1st column [2 3 5] - these are prime
> divisors of |G|. The 2nd column is [3 2 1], this means that the centraliser C_G(1a)
> of the element 1a (i.e. the whole group G) is of order 2^3*3^2*5=360.
> Similraly, the 3rd column is [3 . .] (ie [3 0 0]), meaning that the order of the centraliser C_G(2a)
> of an element in the conjugacy class 2a is 2^3*3^0*5^0=8. So we can also compute the size of
> the conjugacy class 2a, as |G|/|C_G(2a)|=45.

Wonderful and clear explanation I have never seen before. If I
understand correctly, the size of the conjugacy class is the order of
this class, i.e., the number of elements belong to this class. So, in
this case, there are 45 elements in the conjugacy class 2a.

2a    360/ 2^3*3^0*5^0 =360/8=45
3a    360/ 2^0*3^2*5^0 = 360/9=40
3b    360/ 2^0*3^2*5^0 = 360/9=40
4a    360/ 2^2*3^0*5^0 = 360/4=90
5a    360/ 2^0*3^0*5^1 = 360/5=72
5b    360/ 2^0*3^0*5^1 = 360/5=72

Therefore, we have the following relationships:

45 + 40 + 40+ 90+72+72 = 359

359 + 1 (the <identity>) = 360

> Looking at, say, the rows
>        1a 2a 3a 3b 4a 5a 5b
>     ........
>     3P 1a 2a 1a 1a 4a 5b 5a
>
> tells you that the 3rd power of an element in 5a is in 5b.

Got it. Remarkable explanation again.

> HTH
> Dima

Yours,
Hongyi