Inverse of Group Product/General Result/Proof 2

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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 $P \paren n$ be the proposition:

$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$


$P \paren 1$ is (trivially) true, as this just says:

$\paren {a_1}^{-1} = a_1^{-1}$


Basis for the Induction

$P \paren 2$ 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 $P \paren k$ is true, where $k \ge 2$, then it logically follows that $P \paren {k + 1}$ 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:


\(\displaystyle \paren {a_1 \circ a_2 \circ \cdots \circ a_k \circ a_{k + 1} }^{-1}\) \(=\) \(\displaystyle \paren {\paren {a_1 \circ a_2 \circ \cdots \circ a_k} \circ a_{k + 1} }^{-1}\) $\quad$ General Associativity Theorem $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a_{k + 1}^{-1} \circ \paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1}\) $\quad$ Basis for the Induction $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a_{k + 1}^{-1} \circ \paren {a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1} }\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a_{k + 1}^{-1} \circ a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}\) $\quad$ General Associativity Theorem $\quad$

So $P \paren k \implies P \paren {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$

$\blacksquare$


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