Problem

Prove that \((a_1a_2…a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}…a_1^{-1}\) for all \(a_1, a_2,…,a_n \in G\).

Solution

By associativity of \(G\) and because for each \(1 \le i \le n\), \(a_ia_i^{-1} = 1\), for any \(a_1, a_2,…,a_n \in G\),

$$ (a_1a_2…a_n)a_n^{-1}a_{n-1}^{-1}…a_1^{-1} = a_1a_2…(a_na_n^{-1})…a_2^{-1}a_1^{-1} = … = 1 $$ and

$$a_n^{-1}a_{n-1}^{-1}…a_1^{-1}(a_1a_2…a_n) = a_n^{-1}…a_2^{-1}(a_1^{-1}a_1)a_2…a_n = … = 1$$

Therefore, by definition of inverse, \((a_1a_2…a_n)^{-1} = a_n^{-1}…a_1^{-1}\).

$$\tag*{$\blacksquare$}$$