We’ve seen what a group is and a few examples, so I thought I’d show you some useful consequences from our definition. (A lemma is the mathematical name for a small result.)
Lemma 1. The identity element of a group is unique.
Proof. Assume this were not the case, so that we have 2 identity elements e and f. These have that property that g∗e=e∗g=g and g∗f=f∗g=g for all elements g in our group. Now we may consider the product e∗f. Substituting f into e∗g=g we obtain that e∗f=f. But substituting e into g∗f=g we obtain e∗f=e. Thus e∗f=f=e, and there is only one identity element.
Lemma 2. For any given element of a group, the inverse is unique.
Proof. Again, assume this were not the case, so that -for some element a– we have two inverses. Label these a-1 and â-1. Now, consider a-1∗a∗â-1. Using associativity and the definition of an inverse, we have (a-1∗a)∗â-1=â-1 but also that a-1∗(a∗â-1)=a-1. Hence a-1∗a∗â-1=â-1=a-1, meaning that there is only one inverse for any given element.
Lemma 3. Let f be an element such that f∗g=g for some element g. Then f is the identity element of the group.
Proof. Let e denote the identity element of our group. Our assumption says that f∗g=g, for some element g. Since we are working within a group, we have an inverse of g, which we’ll denote g-1. Thus (f∗g)∗g-1=g∗g-1, which by associativity can be rewritten as f∗(g∗g-1)=g∗g-1, i.e. f∗e=e. Now as f∗e=f (by the definition of e, the identity element) we have that f=e.
A similar proof follows if we assume that there is an element f satisfying g∗f=g for some g in the group. Note that Lemma 3 means that the identity is the only idempotent element (an x such that x2=x).
Definition. Let G be a group. If an element e satisfies (i) and (ii) it is called an identity element.
i) g∗e = g for all g in G;
ii) e∗g = g for all g in G.
Furthermore, if an element e satisfies (i) then it is a right identity (it behaves like an identity when you multiply by it on the right). Similarly an element e that satisfies (ii) is called a left identity.
We may use Lemma 3 to see that, in a group, (i) and (ii) cannot occur independently: if one is true then the other must also be true (and so e must be an identity element).
Lemma 4. If ac=bc, then a=b (called the right cancellative law).
Proof. We essentially do what we did in the proof of Lemma 3: multiply both sides on the right by c-1. The proof is left as an exercise.
Again, a similar proof shows that the left cancellative law (that if ca=cb then a=b) holds for every group.
Next we’ll see our second family of finite groups!