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:
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
Now applying the inverse axiom gives
Finally applying the identity axiom leaves us with
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.
operate on each side from the left by a-1:
Now apply the associative axiom
Apply the inverse axiom
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
Proof:
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.
Proof:
The companion theorem to Theorem 3.3 is the left cancellation law,
Theorem 3.3a: If x a = x b then a = b.
Proof:
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
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.
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.
Proof:
Now we can establish the two-sided versions.
Theorem: a-1 a = e.
Proof:
Now we have established that a-1 a multiplied by itself
is itself. thus by the lemma it is e. or
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.
Proof:
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.
Cayley Tables
© 1998 by Arfur Dogfrey
(a b) b-1 = e b-1
a (b b-1) = e b-1
a e = e b-1
a = b-1
(a x) x-1 = (b x) x-1
a (x x-1) = b (x x-1)
a e = b e
a=b
(a a) a-1 = a a-1
a (a a-1) = a a-1
a e = e
a = e
a-1 (a a-1) = a-1 e = a-1
[a-1 (a a-1)] a = a-1 a
[(a-1 a) a-1] a = a-1 a
(a-1 a) (a-1 a) = a-1 a
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