We have been introduced to all the groups of order 8. Three of them are abelian - ℤ8, ℤ4✕ℤ2 and ℤ2✕ℤ2✕ℤ2 - and two are non-abelian, the dihedral 4 group, a.k.a. symmetries of the square, and the quaternions. In an abelian group, every element is in a singleton conjugacy class, so the character table will be 8x8. Our two non-abelian groups both have five conjugacy classes, and since the sum of the squares of the dimensions of the representations must add to 8, there are four one-dimensional irreducible representations and one irreducible representation that is two-dimensional.
Let me begin with ℤ2✕ℤ2✕ℤ2, where the elements will be all the bit strings of length 3 - 000, 001, 010, 011, 100, 101, 110 and 111 - and the group operation will be bitwise exclusive or. 000 is the identity element and every element is its own inverse. As usual, we will fill in the first row with all 1 and likewise the first column.
xor 000 001 010 011 100 101 110 111
1st | 1 1 1 1 1 1 1 1
2nd | 1
3rd | 1
4th | 1
5th | 1
6th | 1
7th | 1
8th | 1
Because every element is its own inverse, the only entries will be 1 and -1. Once it is decided how to map 001, 010 and 100, all the other entries will be determined. Let's start with 001 sent to -1 and the other strings with a single 1 sent to 1.
xor 000 001 010 011 100 101 110 111
1st | 1 1 1 1 1 1 1 1
2nd | 1 -1 1 -1 1 -1 1 -1
3rd | 1
4th | 1
5th | 1
6th | 1
7th | 1
8th | 1
Next, 010 is sent to -1 and the other strings with a single 1 are sent to 1.
xor 000 001 010 011 100 101 110 111
1st | 1 1 1 1 1 1 1 1
2nd | 1 -1 1 -1 1 -1 1 -1
3rd | 1 1 -1 -1 1 1 -1 -1
4th | 1
5th | 1
6th | 1
7th | 1
8th | 1
Next, 100 is sent to -1 and the other strings with a single 1 are sent to 1.
xor 000 001 010 011 100 101 110 111
1st | 1 1 1 1 1 1 1 1
2nd | 1 -1 1 -1 1 -1 1 -1
3rd | 1 1 -1 -1 1 1 -1 -1
4th | 1 1 1 1 -1 -1 -1 -1
5th | 1
6th | 1
7th | 1
8th | 1
For representations 5 through 7, two of the single 1 bit values are sent to -1.
xor 000 001 010 011 100 101 110 111
1st | 1 1 1 1 1 1 1 1
2nd | 1 -1 1 -1 1 -1 1 -1
3rd | 1 1 -1 -1 1 1 -1 -1
4th | 1 1 1 1 -1 -1 -1 -1
5th | 1 -1 -1 1 1 -1 -1 1
6th | 1 -1 1 -1 -1 1 -1 1
7th | 1 1 -1 -1 -1 -1 1 1
8th | 1
For representation 8, all of the single 1 bit values are sent to -1.
xor 000 001 010 011 100 101 110 111
1st | 1 1 1 1 1 1 1 1
2nd | 1 -1 1 -1 1 -1 1 -1
3rd | 1 1 -1 -1 1 1 -1 -1
4th | 1 1 1 1 -1 -1 -1 -1
5th | 1 -1 -1 1 1 -1 -1 1
6th | 1 -1 1 -1 -1 1 -1 1
7th | 1 1 -1 -1 -1 -1 1 1
8th | 1 -1 -1 1 -1 1 1 -1
It is easy to check orthogonality in each case. If we take any two different rows, they will have four entries where the signs disagree, so the dot product will be 4 - 4 = 0. The dot product of any row with itself is 8.
Tomorrow, I will present the character table for ℤ4✕ℤ2.
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