Let $G$ be a group and $a_i \in G$ for all $ n \in N$.
- Prove that $(a_1a_2a_3)^{-1} = a_3^{-1}a_2^{-1}a_1^{-1}$
My Solution: $(a_1a_2a_3)^{-1}$ = $(a_1a_2)^{-1}(a_3)^{-1}$ = $a_1^{-1}a_2^{-1}a_3^{-1}$
(Now as only Abelian groups are commutative then how can the order change in the answer. Either that or I'm missing something.)