If $(G,*)$ is a group and $a_1,a_2,\ldots,a_n \in G$, prove that$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Question

If $(G,*)$ is a group and $a_1,a_2,\ldots,a_n \in G$, prove that $(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Attempt: For the simplify and the convenience, write $a*b=ab$.

I'll prove it by induction on $n$.

Let $a_1,a_2,\ldots,a_n \in G$. Let $P(n)$ be the statement: $(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Base Case: Clearly $P(1)$ is true since $(a_1)^{-1} = a_1^{-1}$. (Is it enough? What if I take $P(1)$ and $P(2)$ for the base case?)

Inductive Step: Assume that $P(k)$ is true. That is, $(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$. To show that $P(k+1)$ is also true, here it is: \begin{align*} (a_1a_2\ldots a_{k+1})^{-1} &= ((a_1a_2\ldots a_k)a_{k+1})^{-1} \\ &= (a_{k+1})^{-1}(a_1a_2\ldots a_k)^{-1} \\ &= a_{k+1}^{-1}(a_k^{-1}a_{k-1}^{-1}\ldots a_1^{-1}) \\ &=a_{k+1}^{-1}a_k^{-1}a_{k-1}^{-1}\ldots a_1^{-1} \end{align*} as desired.

Hence, $(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$ for all $a_1,a_2,\ldots,a_n \in G$. $\Box$

Is the approach above true?

Answer 1

Yes, this looks good with $P(2)$ which you use implicitly in your inductive step, if you're really trying to be exacting in the amount of detail, you may want to be explicit in you use of the base case and the inductive hypothesis in your inductive step, but I think almost anyone who read this would have no problem with it.

Answer 2

Your induction proof looks fine.

Alternatively, you can use uniqueness of inverses and just check that $$ (a_1*\cdots*a_n)(a_n^{-1}*\cdots*a_1^{-1}) = e $$ using associativity.