Let $a,b$ be two elements of a group $G$ such that $a^7 = b^7$ and $a^3 = b^3$ implies $a = b$, then which of the following is possible ?
(i) $a,b \in\Bbb Z_{6}$,
(ii) $a,b \in\Bbb Z_{3}$,
(iii) $a,b \in \Bbb Z \times \Bbb Z$,
(iv) $a,b \in \Bbb Z\times \Bbb Z \times \Bbb Z$.
Now, here all groups are abelian so,
$(ab^{-1})^7 = e$ and $(ab^{-1})^3 = e$
So, we get $o(ab^{-1})\mid 3$ and $o(ab^{-1})\mid 7$, this implies $o(ab^{-1}) = 1$
Hence $a = b$
Since , I have only utilised the property that given Group $G$ is abelian, this should be true for finite as well as infinite groups.
So, all options should be correct
Is my answer and reasoning correct ? Can someone please verify ?
Thank you.
If $a^3=b^3$ then $a^6=b^6$. Now multiply both sides of $a^7=b^7$ by $a^{-6}$ and $b^{-6}$ respectively.