Equivalence of Cayley Tables
How do we determine if two algebras are isomorphic (more about isomorpisms here) be comparing their Cayley tables? It is not good enough to compare the two Cayley tables term by term because if we swap two terms or invert a given term then the algebras will still be isomorphic so we need to do a more complex comparison.
If two multipications 'o' and '*' are isomorpic then:
φ(v1) o φ(v2) = φ(v1 * v2)
or rearanging:
v1 * v2 = φ-1(φ(v1) o φ(v2))
The function between the two groups φ and its inverse can be represented by two columns for example:
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So to apply this to a Cayley table we apply φ() to both the inputs (which are the row and column headings) then apply φ-1() to the output (which are the entries in the body of the table). We may then reorder the rows and columns to put them in the original order.
Two simple examples of isomorphisms are 'inverting a term' and 'swapping two terms' in these cases φ is its own inverse so we can apply this to all the inputs and the outputs.
Invert a term
We can invert a term, say 'k' without changing the group. To do this we need to invert all of:
- the 'k' row
- the 'k' column
- any entry containing 'k'
If 2 of these apply they cancel out and the term doesn't change.
So in the following case we can invert the highlighted terms without changing the group:
a*b |
b.1 | b.i | b.j | b.k |
| a.1 | 1 | i | j | k |
| a.i | i | -1 | k | -j |
| a.j | j | -k => k | -1 | i |
| a.k | k | j | -i | -1 |
Swap terms
We could swap k and j for instance. Where k occurs in the table write j and where j occurs in the table write k:
a*b |
b.1 | b.i | b.k | b.j |
| a.1 | 1 | i | k | j |
| a.i | i | -1 | j | -k |
| a.k | k | -j | -1 | i |
| a.j | j | k | -i | -1 |
Then swap the rows and columns to get them back in the same order:
a*b |
b.1 | b.i | b.j | b.k |
| a.1 | 1 | i | j | k |
| a.i | i | -1 | -k | j |
| a.j | j | k | -1 | -i |
| a.k | k | -j | i | -1 |
