Exercise 1.1.15 ($(a_1a_2\cdots a_n)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1}$)

Question

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

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} We define the proposition $\mathscr{P}(n)$ by
$$\mathscr{P}(n) \iff \forall (a_1,a_2,\ldots,a_n) \in G^n,\ (a_1a_2\cdots a_n)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1}.$$
\begin{itemize}
\item For all $a_1 \in G$, $a_1^{-1} = a_1^{-1}$, so $\mathscr{P}(1)$ is true.

\item Suppose that $\mathscr{P}(n)$ is true for some positive integer $n$. 

Let $(a_1,a_2,\ldots,a_n, a_{n+1}) \in G^{n+1}$. Then
\begin{align*}
(a_1 a_2 \cdots a_n a_{n+1})^{-1} &=a_{n+1}^{-1}(a_1 a_2 \cdots a_n)^{-1} &(	ext{by Proposition 1)}\
&=a_{n+1}^{-1} ( a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1}) & (	ext{by induction hypothesis)}\
&= a_{n+1}^{-1} a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1},
\end{align*}
so $\mathscr{P}(n+1)$ is true.
\item The induction is done, which proves that for all $n \in \Z^+$ and for all $a_1,a_2,\ldots,a_n $ in $G$,
$$(a_1a_2\cdots a_n)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1}.$$
\end{itemize}

\end{proof}

Exercise 1.1.15 ($(a_1a_2\cdots a_n)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1}$)

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Exercise 1.1.15 ($(a_1a_2\cdots a_n)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1}$)

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