Math, asked by swati420, 1 year ago

define a group and show that the cancellation law holds in a group

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Answered by Aditya72779
4
The Cancellation Law for Groups
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The Cancellation Law for Groups
The Cancellation Law for Groups
On the Basic Theorems Regarding Groups page we looked at a whole bunch of theorems. We will now look at another important property of groups in that the cancellation law always applies.

Theorem 1: Let (G,∗) be a group and let a,b,c∈G. If a∗b=a∗c or b∗a=c∗a then b=c.
Proof: Let a−1∈G denote the inverse of a under ∗. Suppose that a∗b=a∗c. Then:
(1)
a∗b=a∗c(a−1∗a)∗b=(a−1∗a)∗ce∗b=e∗cb=c
Similarly, suppose now that b∗a=c∗a. Then:
(2)
b∗a=c∗a(b∗a)∗a−1=(c∗a)∗a−1b∗(a∗a−1)=c∗(a∗a−1)b∗e=c∗eb=c■
It is very important to note that the cancellation law holds with regards to the operation ∗ for any group (G,∗). We will see that the cancellation law does not necessarily hold with respect to an operation on a set when we look at algebraic structures with two defined operations.

It is also important to note that if a∗b=c∗a or b∗a=a∗c then we cannot necessarily deduce that b=c because we would then require the additional property that ∗ is commutative which is not one of the group axioms (but instead one of the Abelian group axioms).
Answered by AlluringNightingale
1

Answer :

Group : An algebraic system (G,*) is said to be a group if the following condition are satisfied :

  1. G is closed under *
  2. G is associative under *
  3. G has a unique identity element
  4. Every element of G has a unique inverse in G

Moreover , if a group (G,*) also holds commutative property , then it is called commutative group or abelian group .

To prove :

If a , b , c are any three arbitrary elements of a group G , then

ab = ac → b = c (left cancellation law)

ba = ca → b = c (right cancellation law)

Proof :

Let a⁻¹ ∈ G be the inverse element of a ∈ G , then we have a⁻¹a = aa⁻¹ = e , where e is the identity element in G .

Now , ab = ac

→ a⁻¹(ab) = a⁻¹(ac)

→ (a⁻¹a)b = (a⁻¹a)c

→ eb = ec

→ b = c

Also , ba = ca

→ (ba)a⁻¹ = (ca)a⁻¹

→ b(aa⁻¹) = c(aa⁻¹)

→ be = ce

→ b = c

Hence proved .

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