Inverse of Group Product/General Result

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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 1

We have that a group is a monoid, all of whose elements are invertible.

The result follows from Inverse of Product/Monoid/General Result.

$\blacksquare$


Proof 2

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:


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