20251106
Traded silence in class for a couple of words, tried miserably to explain what I’m about to write in detail but failed to be clear. Thoughts of adding some cool information were passing, so I tried to add into the mix, only to be asked to repeat, "Excuse me?"
Take a look at any book about algebraic groups you know; three axioms will always appear. They can be presented as logical statements just like in the following list, or the author can spare logical notation by presenting them as identities. They can also be explained using words (why not?), but logic should suffice.
(G1) (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c).
(G2) ∃ e ∈ G such that e ⋅ a = a, ∀a ∈ G.
(G3) ∀ a ∈ G, ∃ b ∈ G such that b ⋅ a = e.
(A) ∃ e ∈ G such that a ⋅ e = a, ∀a ∈ G.
(B) ∀ a ∈ G, ∃ b ∈ G such that a ⋅ b = e.
The book Grupos, Corpos e Teoria de Galois, a favorite of mine, presents the axioms (G1), (G2), and (G3), leaving as an exercise the proof that they are equivalent to the axioms (G1), (A), and (B). That part isn’t difficult to verify and is usually addressed in any standard reference on the topic. The follow-up part of the exercise, however, is slightly more challenging, as we have to prove that (G1), (G2), and (B) (or alternatively (G1), (A), and (G3)) are not enough to define a group. That’s when we have to find or construct a counterexample.
This is only a small detail of the theory, but I appreciate the notion that you could go wrong with the choice of axioms. Also, it is simple enough to serve as a first experiment with math using only pure HTML.
Time to solve the exercise. There’s really no big ideas involved, so the explanation for each step should be evident; just remember that we have the axioms (G1), (A), and (B) at hand.
(G1) ⇔ (G1).
(B) ⇔ ∀ a ∈ G, ∃ b ∈ G such that ab = e.
⟹ eab = ee ⟹ eab = e ⟹ eaba = ea
⟹ e(ab)a = a ⟹ eea = a
⟹ (ee)a = a ⟹ ea = a, ∀a ∈ G ⇔ (G2).
b ∈ G ⟹ ∃ c ∈ G such that bc = e.
ab = e ⟹ bab = b ⟹ babc = bc
⟹ (ba)(bc) = bc ⟹ (ba)e = e
⟹ ba = e, ∀ a ∈ G ⇔ (G3).
Therefore its clear that (G1),(G2),(G3) ⇔ (G1),(A),(B).
To construct our counterexample, lets make use of a Cayley Table. We consider the set A = {a, e} with the operation ⋅ defined as follows:
| ⋅ | a | e |
|---|---|---|
| a | a | e |
| e | a | e |
Those familiar with group theory may have already noticed, just from the table, that A = (A, ⋅) does not form a group, as each element of A should appear only once in each row and column. That said, we’re still going to verify that (G1), (G2), and (B) hold and show that (A, ⋅) is not a group.
Taking our time to cover each case, we see that
(aa)a = a = a(aa),
(aa)e = e = a(ae),
(ae)e = e = a(ee),
(ee)e = e = e(ee),
(ee)a = a = e(ea),
(ea)a = a = e(aa),
(ea)e = e = e(ae),
(ae)a = a = a(ea).
Thus, (G1) holds.
∃ e ∈ G such that ea = a and ee = e, which means that (G2) holds.
ae = e and ee = e, therefore (B) also holds. That is, we have axioms (G1), (G2), and (B) as true.
To prove that A = (A, ⋅) is not a group, you could argue a lot of things, but it’s enough to notice that ea = a ≠ ae; our neutral element does not behave well, and the equivalence between (G2) and (A) is not found.
That's it. Don't mess right and left; start with left (right) identity elements and inverses, or start with both at the same time.