**Theorem 4.4**: If we compose
two invertible isometries, the result is an invertible
isometry.

**Proof**: Let* **M*_{1 }and* **M*_{2 }be the two
isometries. Let* **A *and* B *be two points in
the plane. Then

= |*AB*|

by the definition of an isometry.

Next, suppose that* **M*_{1 }and* **M*_{2 }are invertible, i.e., there exist* **M*_{1}^{-1} and* **M*_{2}^{-1} such that

*M*_{1}*M*_{1}^{-1} = *M*_{1}^{-1}*M*_{1} = *I*

and

* **M*_{2}*M*_{2}^{-1} = *M*_{2}^{-1}*M _{2}* =

where* I *is the identity function. Then

(*M*_{1}*M*_{2})(*M*_{2}^{-1}*M*_{1}^{-1})

= *M*_{1}(*M*_{2}*M*_{2}^{-1})*M*_{1}^{-1}

= *M*_{1}*M*_{1}^{-1} = *I*

and

(*M*_{2}^{-1}*M*_{1}^{-1})(*M*_{1}*M*_{2})

= *M*_{2}^{-1}(*M*_{1}^{-1}*M*_{1})*M*_{2}

= *M*_{2}^{-1}*M*_{2} = *I*

So *M*_{2}^{-1}*M*_{1}^{-1} is an inverse for *M*_{1}*M*_{2}.