Isomorphism between Symmetry Group of Regular Pentagon and Subgroup of Symmetric Group

Theorem
Let $\PP = ABCDE$ denote a regular pentagon.

Let $\struct {\PP, \circ}$ be the symmetry group of $\PP$, where the various symmetry mappings are identified as:
 * the identity mapping $e$
 * the rotations $r, r^2, r^3, r^4$ of $72^\circ, 144^\circ, 216^\circ, 288^\circ$ around the center of $\PP$ anticlockwise respectively
 * the reflections $t_A, t_B, t_C, t_D, t_E$ about the lines through the center of $\PP$ and the vertices $A$ to $E$ respectively.


 * Symmetry-Group-of-Regular-Pentagon.png

Let $\struct {S_5, \circ}$ denote the symmetry group on $5$ letters.

Then $\PP$ is isomorphic to a subgroup of $\struct {S_5, \circ}$ of order $10$.

Proof
Let the $A$, $B$, $C$, $D$ and $E$ of $\PP$ be identified with the integers $1$, $2$, $3$, $4$ and $5$ of $S_5$.

Let each of the symmetry mappings of $\PP$ be identified with permutations of $S_n$ according to where the symmetry mappings moves the vertices of $\PP$.

We express these in two-row notation, which we construct by inspection:

These are now identified with permutations of $S_5$ and hence elements of $\struct {S_5, \circ}$, which we express in cycle notation:

The isomorphism is implicit in the construction.