In the last post, we discovered that for any prime p, ℤp is a finite field, where the entire set under addition is an abelian group and ℤp-{0} is an abelian group under multiplication. On the other hand, if n is a composite number, ℤn-{0} is not a group under multiplication because there will be at least two numbers between 1 and n-1 that multiply to n, which in modulo n arithmetic means j*k = 0, and neither j or k will have multiplicative inverses.
Let us instead look at an additive group of four elements, 0, 1, a and 1+a. Here is the Cayley table for addition.
0 1 a 1+a
0 | 0 1 a 1+a
1 | 1 0 1+a a
a | a 1+a 0 1
1+a | 1+a a 1 0
Every element is its own additive inverse, so this is isomorphic to the Klein four-group. What about the non-zero elements under multiplication? There are three such elements, so if this is a group, it must be isomorphic to ℤ3. Let's figure out the Cayley table.
1 a 1+a
1 | 1 a 1+aa | a ? ?
1+a | 1+a ? ?
Recall that every element must show up in every row and column, so a*(1+a) must equal 1, and likewise (1+a)*a.
1 a 1+a
1 | 1 a 1+aa | a ? 1
1+a | 1+a 1 ?
The other two question marks can be sussed out by using the Latin square rule.
1 a 1+a
1 | 1 a 1+aa | a 1+a 1
1+a | 1+a 1 a
In regular multiplication, (1+a)²= 1 + 2a + a², but because this group uses mod 2 arithmetic, this simplifies to 1 + a². On the table we see a² = 1+a, so we get 1+ 1+a, and mod 2 simplifies this to a. I leave it as an exercise for the reader to see that any combination of these four elements, will be consistent with our addition and multiplication tables.
In the literature, a field whose order is q = pⁿ is called Fq or 𝔽q or GF(q), which stands for Galois field.
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