Cancellation Law in Group Theory: Proof and Validity

[dijkarte]

Let a, b, and c be elements of group <G, *>, then
Can we apply cancellation law here:
a*b = c*a?

I could prove we can cancel a. However, some algebra
Texts do not state this which makes me
Uncertain about the validity of my proof.

 

[wisvuze]

Let a, b, and c be elements of group <G, *>, then
Can we apply cancellation law here:
a*b = c*a?

I could prove we can cancel a. However, some algebra
Texts do not state this which makes me
Uncertain about the validity of my proof.

The cancellation law will work if a*b = a*c. If a*b = c*a, it may not be true that b = c. Instead, a b a^-1 = c ( for example, consider the permutation compositions ( 1 2 ) ( 2 3 1 ) = ( 1 3 2 ) ( 1 2 ) , but ( 2 3 1 ) != ( 1 3 2 ) )

 

[micromass]

It would be a useful exercise for you to find your mistake in your proof by applying your proof to wis' counterexample. Going wrong somewhere is always useful as long as you find out why.

Note that the cancellation law you stat does hold for commutative groups! Maybe your proof uses commutativity somewhere?

 

[dijkarte]

Here's my proof: Assuming a group <G, *>

Let a * b = c * a where a, b, c, all belong to G

Let a' denote inverse of a w.r.t the operation *

Let e denote identity for the operation *

a' * (a * b) = a' * (c * a)

Since an inverse exists for all elements of the group G, we can write a' * a = a * a' = e
So a' can be applied on both left and right.

Hence,

a' * (a * b) = (c * a) * a'

(a' * a) * b = c * (a * a') (applying associativity on both sides)

e * b = c * e ==> b = c

Is this correct?

 

[micromass]

No, it's not correct because wis already gave you a counterexample. Apply wis his counterexample to your proof to see where it went wrong.

 

[SteveL27]

Here's my proof: Assuming a group <G, *>

Let a * b = c * a where a, b, c, all belong to G

Let a' denote inverse of a w.r.t the operation *

Let e denote identity for the operation *

a' * (a * b) = a' * (c * a)

Since an inverse exists for all elements of the group G, we can write a' * a = a * a' = e
So a' can be applied on both left and right.

No, how do you decide you can apply a' both left and right? If a' commutes with every element of G, so does a; and since a was arbitrary, all groups are Abelian. Of course that's not true.

 

[dijkarte]

BY definition of inverse for an operation * on a set, when we say an inverse exists, then we mean the inverse is applicable to both sides of the element. a' * a = a * a' = e.

By definition of a group (not necessarily commutative), an inverse exists such that
a' * a = a * a' = e, for all elements of the group.

 

[micromass]

BY definition of inverse for an operation * on a set, when we say an inverse exists, then we mean the inverse is applicable to both sides of the element. a' * a = a * a' = e.

By definition of a group (not necessarily commutative), an inverse exists such that
a' * a = a * a' = e, for all elements of the group.

Yes, that's true.

But because a*a'=a'*a does NOT mean that a'*b=b*a' for all b!

 

[dijkarte]

I see my mistake, thanks a lot for all. :)