Question

If G is a group such that am bm for the three consecutive integers m for all a,b in G. Then show that G is an abelian group

Answer 1

Answer:

Then prove that G is an abelian group. Proof: Let a and b be any two elements of G. Suppose n, n+1 , n + e are three consecutive integers such that (ab)n =…

Answer 2

Step-by-step explanation:

Let j=i+1,k=i+2 for some i∈Z.

Then we have that (ab)^i=a^ib^i, (ab)^j=a^jb^j and (ab)^k=a^kb^k.

If (ab)^k=a^kb^k, then a^jb^jab=a^jab^jb.

We cancel on the left and right and we have b^ja=ab^j, that is b^iba=ab^j.

Multiply both sides by ai on the left and we get a^ib^iba=a^jb^j, so (ab)^iba=(ab)^j.

But that is (ab)^iba=(ab)^iab.

Cancelling on the left yields ab=ba, which holds for all a,b∈G, and therefore, G is abelian.