Jim Caprioli at The Finland Station

Exercise: Calculate the character table of a non-abelian group ( of a group you are not familiar with )
a) Construct the group
b) Determine the conjugacy classes
c) Fill in the character table sofar
d) Complete the table
Since this is merely a hobby I invent my own exercises :-)

Group Z4 : Z4, the semi-direct product of Z4 and Z4. Call them G and H.
Recall the construction of the semi-direct product.
In order to construct this group a mapping between G and the automorphism group of H must be defined.
The automorphism group of H consists of two automorphisms: I, the identity map and F, the inversion map. I and F form a group isomorphic to C2 since F*F=I.
The morphism p is used in the operation of the new group. p: Z4->Aut(Z4)
p(0) = I
p(1) = F
p(2) = I
p(3) =F
This is clearly a homomorphism.
Using this the group K can be constructed as follows.
The elements are in the set { (g, h) | g in G, h in H } - first elements are { (0,0), (0,1), (0,2), (0,3), (1,1) ... (3,3) }
Group Operation *
( g1, h1 ) * ( g2, h2 ) = ( g1 + g2, h1 + p(g1)[h2] )
( See Planet Math: semi-direct product )

Implemented in Mathematica, for further analysis, the Group has the following definition:

z := Table[{IntegerPart[x/4], Mod[x, 4]}, {x, 0, 15}]
G := FormGroupoid[z,
{Mod[#1[[1]] + #2[[1]], 4],
If[#1[[1]] == 1 || #1[[1]] == 3,
If[#2[[2]] == 1 || #2[[2]] == 3,
Mod[#1[[2]] + #2[[2]] + 2, 4],
Mod[#1[[2]] + #2[[2]], 4]],
Mod[#1[[2]] + #2[[2]], 4]]} &]
( I don't like this code at all but it works. The exercise is not about Mathematica programming which is a science all by itself. )

Most important. Is it indeed a group and above all: is it Non-Abelian??
GroupQ[G]
AbelianQ[G]
>
. True
. False

Indeed. Furthermore I would expect proper subgroups of orders 8, 4, 2. I have found several of them. To the point the conjugacyclasses. Mathematica reports 10 conjugacy classes.

{
{{0, 0}},
{{0, 2}},
{{2, 0}},
{{2, 2}},
{{0, 1}, {0, 3}},
{{1, 0}, {1, 2}},
{{1, 1}, {1, 3}},
{{2, 1}, {2, 3}},
{{3, 0}, {3, 2}},
{{3, 1}, {3, 3}}
}

The conjugacy classes will form the columns of the character table. 4 classes have 1 element. 6 classes have 2 elements. The CT is square so there are 10 representations as well. The sum of the squares of the characters of the identity is equal to the order of the group. In this case 16, so my wildest guess is that we have 8 representations of dimension 1 and 2 representations of dimension 2.

So what is left (for tomorrow)? Calculating the representations, I suppose.