Algebra Solutions: 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 $a_i \in G$ .

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

$(a_1 \cdot \ldots \cdot a_n \cdot a_{n+1})^{-1}$	=	$\left( (a_1 \cdot \ldots \cdot a_n) \cdot a_{n+1} \right)^{-1}$
	=	$a_{n+1}^{-1} \cdot (a_1 \cdot \ldots \cdot a_n)^{-1}$
	=	$a_{n+1}^{-1} \cdot a_n^{-1} \cdot \ldots \cdot a_1^{-1}$ ,

using associativity and the base case where necessary.
Thus by induction, $(a_1 \cdot \ldots \cdot a_n)^{-1} = a_n^{-1} \cdot \ldots \cdot a_1^{-1}$ for all $n \in \mathbb{Z}^+$ and $a_i \in G$ .

Algebra Solutions

The inverse of a product is the reversed product of inverses

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