Symmetry of an equilateral triangle

Question

I am currently studying "Elements of Abstract Algebra" by Allan Clark. In one of the exercises(26η) of the book he asks: "Show that a symmetry of an equilateral triangle ABC is completely determined by the way it transforms the vertices."

I do not understand what exactly I need to show.

Thanks , in advance

Answer 1

You have to proof two things:

1) given two symmetries $f,g$ of the triangle $ABC$, if $f(A)=g(A)$, $f(B)=g(B)$, and $f(C)=g(C)$, then $f$ and $g$ are the same symmetry.

and

2) given any permutation of the vertices $A,B,C$ there exists a symmetry of the triangle that permutes the vertices in the given way.

Answer 2

You can show that a symmetry that has $3$ non-linear fixed points is the identity

In order to do that name $A,B,C$ these points and show that the symmetry will preserve the sides of the triangle $ABC$.Then for an arbitrary point $M$ one of the lines $AM,BM,CM$ will have a common point with a side of the triangle,so this line will also be preserved by the symmetry,hence $f(M)=M$

Now for any two symmetries $f,g$ that are:$f(A)=g(A),f(B)=g(B),f(C)=g(C)$ for the $f\circ g^{-1}$ you will have the above property so $f\circ g^{-1}=id\Rightarrow f=g$