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

Proof: Let   M1   and   M2   be the two isometries. Let   A   and   B   be two points in the plane. Then

|M2(M1(A)M2(M1(B)|

= |M1(A)M1(B)|

= |AB|

by the definition of an isometry.

Next, suppose that   M1   and   M2   are invertible, i.e., there exist   M1-1   and   M2-1  such that

M1M1-1 = M1-1M1 = I

and

M2M2-1 = M2-1M2 =

I

where   I   is the identity function. Then

(M1M2)(M2-1M1-1)

= M1(M2M2-1)M1-1

= M1M1-1 = I

and

(M2-1M1-1)(M1M2)

= M2-1(M1-1M1)M2

= M2-1M2 = I

So M2-1M1-1 is an inverse for M1M2.

next theorem (4.5)