If you find any mistakes, please make a comment! Thank you.

The inverse of a product is the reversed product of inverses


Let $G$ be a group. Prove that $$(a_1 \cdot \ldots \cdot a_n)^{-1} = a_n^{-1} \cdot \ldots \cdot a_1^{-1}$$ for all $n \in \mathbb{Z}^+$ and $a_i \in G$.


Solution: For $n = 1$, note that for all $a_1 \in G$ we have $a_1^{-1} = a_1^{-1}$.

Now for $n \geq 2$ we proceed by induction on $n$. For the base case, note that for all $a_1, a_2 \in G$ we have $$(a_1 \cdot a_2)^{-1} = a_2^{-1} \cdot a_1^{-1}$$ since $$a_1 \cdot a_2 \cdot a_2^{-1} a_1^{-1} = 1.$$ For the inductive step, suppose that for some $n \geq 2$, for all $a_i \in G$ we have $$(a_1 \cdot \ldots \cdot a_n)^{-1} = a_n^{-1} \cdot \ldots \cdot a_1^{-1}.$$ Then given some $a_{n+1} \in G$, we have \begin{align*} (a_1 \cdot \ldots \cdot a_n \cdot a_{n+1})^{-1} = &\ \left( (a_1 \cdot \ldots \cdot a_n) \cdot a_{n+1} \right)^{-1}\\ = &\ a_{n+1}^{-1} \cdot (a_1 \cdot \ldots \cdot a_n)^{-1}\\ = &\ a_{n+1}^{-1} \cdot a_n^{-1} \cdot \ldots \cdot a_1^{-1}, \end{align*} using associativity and the base case where necessary.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu