Group Housekeeping Theorems

The first step in the formal theory of groups is to take care of some details. In this lesson we'll prove some basic properties of groups that we will use later on.

Theorem 3.1: The identity element of a group is unique.

First off we must translate this ambiguous English sentence into a well-formed mathematical statement. There are two form of unambiguous mathematical statement that we can transform the sentence into. The first we'll call the weak form. It states:

If an element of the group a has the property that for all elements of the group x it is true that a • x = x • a = x then a a is the identity element.

The strong form says:

If an element of the group a has the property that for some other element x of the group it is true that a • x = x • a = x, then a is the identity element.

The weak form says that there is only one element that always acts like the identity element, namely the identity element. The strong form says that there is only one element that ever acts like the identity element even once.

Note that the strong form of the theorem includes the weak form as a special case but the weak form doesn't include the strong form. Thus it is necessary to prove only the strong form to establish both forms. In the interest of verbosity we will prove first the weak form then the strong form.

Proof of weak form: If a member p of the group has the property that for any x in the group, p • x = x • p = x, then using the group identity, e for x we get

p • e = e

Using the identity behavior of p this is equal to e. But using the identity property of e it is also equal to p.

p • e = p

Thus p and e are equal to the same thing and therefore--

p = e

Proof of strong form: If for one member a of the group it is true that p • a = a then we may multiply both sides of this equation by a-1:

(p • a) • a-1 = a • a-1

This step is valid because of both closure which says that the multiplication exists and the inverses axiom insures that a-1 exists. The next step is to apply the associative axiom to the left side and also apply the inverse axiom to the right side. This yields

p • (a • a-1) = e

Now applying the inverse axiom gives

p • e = e

Finally applying the identity axiom leaves us with

p = e

Now we could have proved the strong form first and killed two theorems with one proof. What reason could there be for proving both forms separately? One reason is that we get more experience proving things from the group axioms and thus get a better feel for the beginnings of group theory. The other, deeper reason has to do with what was used in the two proofs. The proof of the weak form did not use the associative axiom or the inverse axiom. Thus we have established this theorem for any system that has an identity and obeys the closure axiom whether it also obeys the other group axioms or not. The proof of the strong form used all four of the group axioms and so has been established only for systems that obey all four of the group axioms.

Theorem 3.2: If a and b are elements of a group and a • b = e, then b = a-1

This theorem states that there is only one element of a group that acts like the inverse of a namely a-1. The proof is simple, we begin with the equation a • b = e and use group axioms to derive b = a-1.

a • b = e (given)

operate on each side from the left by a-1:

a-1 • (a • b) = a-1 • e

Now apply the associative axiom

(a-1 • a) • b = a-1 • e

Apply the inverse axiom

e • b = a-1 • e
FInally apply the identity axiom to each side,

b = a-1

From now on we will assume that the reader is familiar with the direct applications of the axioms. Such proofs will be written in a shorthand form. For example we will prove the "mirror" version of theorem 3.2 in shorthand form

Theorem 3.2a: If a and b are elements of a group and a • b = e then a = b-1


a • b = e
(a • b) • b
-1 = e • b-1
a • (b • b-1) = e • b-1
a • e = e • b
a = b

Theorem 3.3: If a • x = b • x then a = b.

This is the right cancellation law, we "cancel" the x from the right on both sides of the equation.


a • x = b • x
(a • x) • x
-1 = (b • x) • x-1
a • (x • x
-1) = b • (x • x-1)
a • e = b • e

The companion theorem to Theorem 3.3 is the left cancellation law,

Theorem 3.3a: If x • a = x • b then a = b.


We'll leave the proof of this one as an exercise for the student.

Theorem 3.4: The inverse of a • b is b-1 • a-1.

Note the change of order. This theorem states that the inverse of the product of two elements is the product of the inverses in reverse order.

Proof: Now we can use theorem 3.2. All we have to do is show that b-1 • a-1 acts as the inverse to a • b and Theorem 3.2 guarantees that it is the inverse of a • b

(a • b) • (b-1 • a-1) = a • [b • (b-1 • a-1)] = a • [(b • b-1) • a-1] = a • (e • a-1) = a • a-1 = e

The proof consisted of repeated use of the associative property together with the inverse and identity properties.

By repeated application of Theorem 3.4 we can conclude that (a • b • c)-1 = c-1 • b-1 • a-1 or (a • b • c • d)-1 = d-1 • c-1 • b-1 • a-1 etc.

Again using Theorem 3.2 we can quickly establish our final housekeeping result.

Theorem 3.5: (a-1)-1 = a and e-1 = e

The proof is simply to apply Theorem 3.2 to the equations a-1 • a = e and e • e = e.

Weakening the Axioms

It turns out that (ITOT) the axioms we gave for groups are redundant. That is they explicitly state more than they need in order to characterize group property. Specifically the Identity axiom states that e is both a right identity and a left identity. Similarly the Inverse axiom states that a-1 is both a right inverse of a and a left inverse of a. These axioms could have been sneakier and not given so much information directly. For instance let's replace the Identity axiom and the inverse axiom with the following weaker axioms.

(W Id A): There is an element e such that for any a of the group a • e = a.

(W In A): For every element a of the group there is an element a-1 such that a • a-1 = e.

To show that these weaker, one-sided axioms are just as strong as the original two-sided axioms we'll derive the two-sided form from them.

We'll begin with a preliminary result (a lemma) that will be used in the proofs. Remember, in all that follows we are trying to derive the two-sided axioms from the one-sided ones. Thus we are not allowed to use the usual two-sided axioms in our proofs.

Lemma: If a • a = a then a = e.


a • a = a
(a • a) • a
-1 = a • a-1
a • (a • a
-1) = a • a-1
a • e = e
a = e

Now we can establish the two-sided versions.

Theorem: a-1 • a = e.


a • a-1 = e
-1 • (a • a-1) = a-1 • e = a-1
-1 • (a • a-1)] • a = a-1 • a
-1 • a) • a-1] • a = a-1 • a
-1 • a) • (a-1 • a) = a-1 • a

Now we have established that a-1 • a multiplied by itself is itself. thus by the lemma it is e. or

a-1 • a = e

We have established that if a-1 is a right inverse then it is also a left inverse. Now it remains to show that e is a left identity.

Theorem: For all a in the group e • a = a.


e • a = (a • a-1) • a = a • (a-1 • a) = a • e = a

All of the original two-sided properties have been recovered. We also could have written the weakened axioms from the other side with left identity and left inverses and then proceeded to recover the right inverse and right identity properties. An interesting problem for the student to amuse herself with is to try to recover the two-sided properties from a right-sided identity axiom and a left-sided inverse axiom.

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© 1998 by Arfur Dogfrey