Does the set of all symmetries of a plane figure form a group under composition of functions? Prove that the set of all symmetries of a plane figure form a group under composition of functions.
So I know I need to show the group axioms to be true under composition for symmetries but I'm having trouble doing this. This is what I have so far
Let $P$ be a plane figure.
Let $S_P = \{ S_1, S_2,...\}$ be the set of all symmetries of $P$.
Let $∘$ denote composition of mappings.
Closure: $S_i ∘ S_j (P)$ is a symmetry $S_k \in S_P$
Associativity: $S_i ∘ (S_j ∘ S_k) (P) = (S_i ∘ S_j) ∘ S_k (P)$ as composition is associative.
Identity: $S_e(P) ∘ S_j(P) = S_j(P) ∘ S_e(P) = S_j(P)$ where $S_e (P) = P$
Inverse: $S_i ∘ S_j(P)$ such that $S_i ∘ S_j(P) = S_j ∘ S_i(P) = P$, where $S_i = S_j ^{-1}$.
Any help with this is appreciated.
 A: Recall that a symmetry of a plane figure $P \subseteq \mathbb{R}^{2}$ is an isometry $f : \mathbb{R}^{2} \to \mathbb{R}^{2}$ whereby $f(P) = P$.  Letting $d$ denote the Euclidean metric on $\mathbb{R}^{2}$, recall that a mapping $f : \mathbb{R}^{2} \to \mathbb{R}^{2}$ is an isometry if $$d(f(\vec{v}), f(\vec{w})) = d(\vec{v}, \vec{w})$$ for all $\vec{v}, \vec{w} \in \mathbb{R}^{2}$.
To prove that the set of all symmetries of a plane figure form a group under composition of functions, you need to use the above definition of a symmetry of a plane figure.
Letting $S = S_{P}$ denote the set of all symmetries of a plane figure $P$, let $f, g \in S$ be arbitrary. Since $f : \mathbb{R}^{2} \to \mathbb{R}^{2}$ and $g : \mathbb{R}^{2} \to \mathbb{R}^{2}$ are isometries, we have that $$d(fg(\vec{v}),fg(\vec{w}))=d(g(\vec{v}),g(\vec{w}))=d(\vec{v},\vec{w})$$ for all $\vec{v}, \vec{w} \in \mathbb{R}^{2}$, and we thus have that the composition $fg = f \circ g : \mathbb{R}^{2} \to \mathbb{R}^{2}$ is an isometry. Since $f$ and $g$ are symmetries of $P$, we have that $$fg(P) = f(P) = P$$ and we thus have that $fg$ is also a symmetry of $P$. Therefore, the composition operation $\circ$ is a binary operation on $S$. 
Since the composition of functions is (in general) associative, we have that $\circ$ is an associative binary operation on $S$.  
It is obvious that the identity mapping $\text{id} : \mathbb{R}^{2} \to \mathbb{R}^{2}$ on $\mathbb{R}^{2}$ is an isometry which fixes $P$, and it is obvious that $\text{id} \circ f = f \circ \text{id}$ for all $f \in S$. 
So it remains to prove that the inverse axiom holds. To prove this, prove that each symmetry $f \in S$ is bijective and thus has an inverse $f^{-1}$, and then use the definition of the term symmetry given above to prove that $f^{-1} \in S$ for $f \in S$.  
Reference: "How to Count: an Introduction to Combinatorics", R. Allenby and A. Slomson.
