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.

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Comments

  • 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

    Also, if you’d rather I did not make comments about your minor typos let me know.

    • nbloomf  On October 2, 2010 at 9:41 am

      Actually I really appreciate it. I might not find them otherwise.

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