A group is a set G with three operations:
For practice in presenting basic proofs we'll substitute an alternative set of axioms:
A group is a set G with two operations:
Now to show that this is an acceptable substitute, we just have to show that we can prove a), b), and c) by using i) and ii).
Lemma 1. If s and t are in G, then s * s-1 = t * t-1.
Proof. s * s-1 = s * s-1 * t * t-1 = t * t-1, where we first used the second equality in ii), and then the first.QED
Definition. If s is in G, then we define the element e in G by e = s * s-1.
Lemma 2. The nullary operation "select e" is well defined.
Proof. This amounts to verifying that e is independent of the element s in its definition. But this is the content of Lemma 1. QED
Lemma 3. The element e satisfies a * e = a = e * a for any element a in G.
Proof. Take s in G. Then, by definition, a * e = a * s * s-1. By Axiom ii), a * s * s-1 = a = s * s-1 * a = e * a. QED
Lemma 3.5. If a,b, and x are in G, and a * x = b* x, then a = b.
Proof. Since a * x = b * x, a * x * x-1 = b* x * x-1, and so a = b. QED
Lemma 4. For any x in G, x * x-1 = x-1 * x = e.
Proof. The easy part is x * x-1 = e. In fact this is just Lemma 1. To prove that x-1 * x = e, we need Lemma 3.5. Since x * x-1 = e, we may write x-1 * x * x-1 = x-1 * e = x-1 = e * x-1. Now apply Lemma 3.5 to x-1 * x * x-1 = e * x-1 to get x-1 * x = e. QED
So we've proven that a), b), and c) hold if i) and ii) do. As an exercise, you may prove that i) and ii) hold whenever a), b), and c) do. We'll now continue with our "invisible ink" proofs of basic facts about groups.
Lemma 5. If a,b, and x are in G, and x * a = x * b, then a = b.
Proof. Since x * a = x * b, x-1 * x * a = x-1 * x * b, and so a = b.QED
Lemma 6. If a is in G, then (a-1)-1 = a.
Proof. By definition, (a-1)-1 * a-1 = e. Therefore, (a-1)-1 * a-1 * a = e * a. So (a-1)-1 = a.QED
Lemma 7. If an element z in G satisfies a * z = a or a = z * a for every a in G, then z = e.
Proof. By assumption, e * z = e or e = z * e. But by Lemma 3, z * e = z = e * z. Thus z = e. QED
Lemma 8. If an element z in G satisfies a * z = a or a = z * a for any one element a in G, then z = e.
Proof. Since a * z = a, a-1 * a * z = a-1 * a. So e * z = e. But e * z = z. So z = e. The proof for the other 'side' is the same. QED
Comment. Of course we could have proven Lemma 7 easily after Lemma 8, since Lemma 7 wasn't used to prove Lemma 8. In fact all our Lemmata may be reordered in several ways. There is probably no one best way to do this, but the student is encouraged to "make your own intro". Check your proofs carefully!
Lemma 3.5 and Lemma 5 are called the Right Cancellation Law and the Left Cancellation Law, respectively. It's tempting to think that a set with an associative binary operation which satisfies both of these Laws must be a group, but it's not true. For example, consider the set of all even positive integers with the operation of multiplication. However, if we assume the set is finite, then the Cancellation Laws imply the group axioms.
Proposition 9. If G is a finite set with an associative binary operation * , such that a * x = a * y implies x = y , and x * a = y * a implies x = y, then G is a group.
Proof. We will show that right cancellation implies right solubility (and, by analogy, left cancellation implies left solubility). Since the set G = {x1,x2,...,xn} is finite, any injective mapping from G to G is bijective. For any a in G, the Cancellation Law is the statement that fa: G -> G given by fa(x) = a * x is injective. The statement of right solubility, that every equation a * x = b has a solution for a and b in G is the same as the surjectivity of fa. Clearly the same argument works for left operations.
Now assume G is a set with associative operation *, such that, for a and b in G the equations a * x = b and y * a = b both have solutions x , y in G (not necessarily distinct). .QED