Wheeler, Mechanics

Finally, since b must have an inverse, and its inverse cannot be a, we must fill in the final spot with the
identity, thereby making b its own inverse:
◦ a b
a a b
b b a
Comparing to the boolean table, we see that a simple renaming, a → 1, b → −1 reproduces the boolean
group. The relationship between different representations of a given group is made precise by the idea of an
isomorphism.
Let G = (S, ⊕) and H = (T, ⊗) be two groups and let φ be a one-to-one, onto mapping, φ, between G
and H. Then φ is an isomorphism if it preserves the group product in the sense that for all g 1 , g 2 in G,
φ (g 1 ) ⊗ φ (g 2 ) = φ (g 1 ⊕ g 2 ) (1)
When there exists an isomporphism between G and H, then G and H are said to be isomorphic to one
another.
The definition essentially means that φ provides a renaming of the elements of the group. Thus, suppose
g 1 ⊕ g 2 = g 3 . Thinking of h = φ (g) as the new name for g, and setting
eq.(1) becomes
h 1 = φ (g 1 )
h 2 = φ (g 2 )
h 3 = φ (g 3 )
h 1 ⊗ h 2 = h 3
Applying the group product may be done before or after applying φ with the same result. In the Boolean
case, for example, setting φ (a) = 0 and φ (b) = 1 shows that G = (◦, {a, b}) and H = (⊕ 2 , {0, 1}) are
isomorphic.
Now consider a slightly bigger group. We may find all groups with three elements as follows. Let
G = {{a, b, c} , ⊗} , where the group operation, ⊗, remains to be defined by its multiplication table. In order
for G to be a group, one of the elements must be the identity. Without loss of generality, we pick a = e.
Then the multiplication table becomes
⊗ e b c
e e b c
b b
c c
Next, we show that no element of the group may occur more than once in any given row or column. To
prove this, suppose some element, z, occurs in the c column twice. Then there are two distinct elements
(say, for generality, x and y) such that
From this we may write
x ⊗ c = z
y ⊗ c = z
x ⊗ c = y ⊗ c
But since c must be invertible, we may multiply both sides of this equation by c −1 and use associativity:
(x ⊗ c) ⊗ c −1 = (y ⊗ c) ⊗ c −1
x ⊗ ( c ⊗ c −1) = y ⊗ ( c ⊗ c −1)
x ⊗ e = y ⊗ e
x = y
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