Defining a non-abelian finite group using generators

 It's easiest to understand a finite group by having the Cayley table handy, or a list of all the matrices, but the quickest shorthand is to give the generators and the relationships between them. Here is an example.

a⁴=1, b²=1, ab = ba³

 

When listing generators like a⁴=1, that means a raised to the fourth power is the is the lowest positive power of a to equal the identity. The third statement tells us the group is non-abelian, but we haven't been told what ba is equal to. Here is how we find it.

ab = ba³ multiply by b in front of both sides of the equality.

bab = b²a³  Since b²=1, we can remove it from the equation.

bab = a³ Now multiply by b in back of both sides of the equality.

bab² = a³b  Again we can remove b².

ba = a³b

There is no need for any other letters and we can try to fill in the Cayley table as follows.

This is similar to solving a puzzle like Sudoku. In point of fact, both Sudoku puzzles and Cayley tables are Latin squares, nxn matrices with only n elements, with one of each element in every row and every column. When we start the process, we will use the rules of group theory, but at the end of the process, we may be able to fill in entries using the rules of Latin squares.

Because the first row and first column are just multiplication by the identity, they are easy to fill in. Pay no attention to the red and green underlines, they are artifacts from Microsoft Word.

The next easiest step is the powers of a, since any element always commutes with itself and its powers.

If we stipulate that the table shows us the value when the leftmost column entry is the first multiple and the topmost row is the second, then the first four rows are easily done. 

 

 

 

 

 Another easy step is b²=1.

 

 

 

 

 

 

 

 

 

We did the work earlier to show 

ba = a³b, which gives us the entry in the fifth row, second column. The rest of the second column is a³b multiplied in front by a, a² and a³ respectively.

 

 

 

Now we need the canonical form of ba². Change it to (ba)a, which equals a³ba. Now we change this to a³(ba), which gives us a³a³b, or a⁶b. Because a⁴=1, this can b re-written as a²b.

The rest of the row is filled in the same way we filled in the last three entries of the previous column.

The rest of the fourth column must be multiples of b and powers of a. We can use Latin square rules to figure out the missing entries of rows five through eight.

 

 

The bottom quadrant has to be filled with the powers of and the identity, by the rules of Latin squares. I leave it as an exercise for the reader to do these last steps. 

 

And now a not completely surprising reveal, this Cayley table can be mapped to our previous Cayley table of the symmetries of the square, where is a R90° and b is any of the four mirror moves. This points out a dichotomy in group theory. It is completely possible to look at the topic in abstract terms, but it is also possible to explain groups in terms of geometry. There are groups that can be defined in just a few symbols whose definition I "know", but trying to visualize groups geometrically sends me down a very odd rabbit hole indeed.

 

John Von Neumann, who many consider the greatest mathematician of the 20th Century, has a fitting quote for this.

 

"In mathematics, you don't understand things. You just get used to them."