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If G is a group such that a2 = e, for all a ∈ G, then G is
- a)abelian group
- b)non-abelian group
- c)ring
- d)field
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If G is a group such that a2 = e, for all a ∈G, then G isa)abelia...
If G is a group such that a^2 = e for all a in G, then G is an abelian group.
Proof:
Let a, b be arbitrary elements in G. We want to show that ab = ba.
Consider the element (ab)^2. Using the given property, we have (ab)^2 = e. Expanding, we get (ab)(ab) = e. Rearranging, we have a(ba)b = e.
Now, multiplying both sides of the equation on the left by a^(-1) (the inverse of a), we obtain (a^(-1)ab)(ba)b = a^(-1)e. Simplifying, we get eb = a^(-1)e. Since e is the identity element, we have b = a^(-1)e = a^(-1).
Similarly, multiplying the original equation on the right by b^(-1) (the inverse of b), we get a(ba)b(b^(-1)) = e(b^(-1)). Simplifying, we get aea = b^(-1)e. Again, using the fact that e is the identity element, we have a = b^(-1)e = b^(-1).
Therefore, we have a = b^(-1) and b = a^(-1), which implies that ab = ba. Hence, G is an abelian group.
Proof:
Let a, b be arbitrary elements in G. We want to show that ab = ba.
Consider the element (ab)^2. Using the given property, we have (ab)^2 = e. Expanding, we get (ab)(ab) = e. Rearranging, we have a(ba)b = e.
Now, multiplying both sides of the equation on the left by a^(-1) (the inverse of a), we obtain (a^(-1)ab)(ba)b = a^(-1)e. Simplifying, we get eb = a^(-1)e. Since e is the identity element, we have b = a^(-1)e = a^(-1).
Similarly, multiplying the original equation on the right by b^(-1) (the inverse of b), we get a(ba)b(b^(-1)) = e(b^(-1)). Simplifying, we get aea = b^(-1)e. Again, using the fact that e is the identity element, we have a = b^(-1)e = b^(-1).
Therefore, we have a = b^(-1) and b = a^(-1), which implies that ab = ba. Hence, G is an abelian group.
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Community Answer
If G is a group such that a2 = e, for all a ∈G, then G isa)abelia...
A^2 =e for all a €G which means every element of the group has self inverse... and now if any two elements say a, b has order 2 , it is clear ab is also of order 2 (since o(ab) =lcm (o(a), o(b)) ).. so (ab) ^2=e which can be written as e=e *e =a^2 ×b^2 which means abab =aabb and now since a^-1 and b^-1 exists, then composing with these from left and right u can get ab=ba for any a, b €G .. hence u are done
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