2.8. Let a, b be elements of a group G. Assume that a has order 5 and a3b = ba3. Prove that ab = ba.
We have a5=ea^5=ea5=e and a3b=ba3a^3b=ba^3a3b=ba3. Applying a3a^3a3 to the second equation we obtain a3(a3b)=a3(ba3)=(a3b)a3=(ba3)a3a^3(a^3b)=a^3(ba^3)=(a^3b)a^3=(ba^3)a^3a3(a3b)=a3(ba3)=(a3b)a3=(ba3)a3 and hence a6b=ba6a^6b=ba^6a6b=ba6.
As a6=a5⋅a=ea=aa^6=a^5\cdot a=ea=aa6=a5⋅a=ea=a we obtain ab=baab=baab=ba.
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