Inverse of Group Product/General Result/Proof 2
Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
- $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
Proof
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\map P 1$ is (trivially) true, as this just says:
- $\paren {a_1}^{-1} = a_1^{-1}$
Basis for the Induction
$\map P 2$ is the case:
- $\paren {a_1 \circ a_2}^{-1} = a_2^{-1} \circ a_1^{-1}$
which has been proved in Inverse of Group Product.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
Then we need to show:
- $\paren {a_1 \circ a_2 \circ \cdots \circ a_{k + 1} }^{-1} = a_{k + 1}^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
Induction Step
This is our induction step:
\(\ds \paren {a_1 \circ a_2 \circ \cdots \circ a_k \circ a_{k + 1} }^{-1}\) | \(=\) | \(\ds \paren {\paren {a_1 \circ a_2 \circ \cdots \circ a_k} \circ a_{k + 1} }^{-1}\) | General Associativity Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds a_{k + 1}^{-1} \circ \paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1}\) | Basis for the Induction | |||||||||||
\(\ds \) | \(=\) | \(\ds a_{k + 1}^{-1} \circ \paren {a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds a_{k + 1}^{-1} \circ a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}\) | General Associativity Theorem |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$: Theorem $2$