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*. Now, consider

^{-1}*a*

^{-1}*∗a∗*

*â*. Using associativity and the definition of an inverse, we have (

^{-1}*a*

^{-1}*∗a*)∗

*â*=

^{-1}*â*but also that

^{-1}*a*∗(

^{-1}*a∗*

*â*)=

^{-1}*a*. Hence

^{-1}*a*

^{-1}*∗a∗*

*â*=

^{-1}*â*=

^{-1}*a*, meaning that there is only one inverse for any given element.

^{-1}**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∗*which by associativity can be rewritten as

*g*^{-1},*f*∗(

*g∗*)=

*g*^{-1}*g∗*, i.e.

*g*^{-1}*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 *x*^{2}=*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!

a=b in statement of Lemma 4 🙂 Nice blog, thanks again for your input and ideas at the workshop.

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Thank you, made the change! And thank you for presenting your ideas at the workshop.

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[…] Some neat little proofs […]

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