Prove that every cyclic group is abelian.
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Answer:
PROOF:---
Let G be a cyclic group with a generator g∈G.
Namely, we have G=⟨g⟩ (every element in G is some power of g.)
Let a and b be arbitrary elements in G.
Then there exists n,m∈Z such that a=gn and b=gm.
It follows that
ab=gngm=gn+m=gmgn=ba.
Hence we obtain ab=ba for arbitrary a,b∈G.
Thus G is an abelian group.
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Note :
- Group : An algebraic system (G,*) is said to be a group if the following condition are satisfied :
- G is closed under *
- G is associative under *
- G has a unique identity element
- Every element of G has a unique inverse in G
- Abelian group : If a group (G,*) also holds commutative property , then it is called commutative group or abelian group , ie . if x*y = y*x ∀ x , y ∈ (G,*) , then the group G is said to be abelian .
Solution :
To prove :
Every cyclic group is an abelian group .
Proof :
Let G = < a > be a cyclic group with generator a .
Let x and y be any two elements of G . Then there exist integers r and s such that x = aʳ and y = aˢ.
Now ,
→ xy = aʳaˢ
→ xy = aʳ⁺ˢ
→ xy = aˢ⁺ʳ
→ xy = aˢaʳ
→ xy = yx
Thus , we have xy = yx ∀ x , y ∈ G .
→ G is an abelian group .
Hence ,
Hence ,Every cyclic group is an abelian group .
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