## 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 $a_i \in G$.

• r.d.  On October 2, 2010 at 9:28 am

Hey, there’s a few typos with your LaTeX here, or it is not being called to display properly.

• nbloomf  On October 2, 2010 at 9:31 am

Fixed. Thanks!

• r.d.  On October 2, 2010 at 9:37 am