The Group of Symmetries of the Equilateral Triangle

Consider an equilateral triangle whose vertices are labelled points:

Consider a point fixed in the center of this triangle. There are two types of symmetries we can look at. The first is counterclockwise rotational symmetries. We can rotate the triangle by $0^{\circ}$ (or equivalently $360^{\circ}$), $120^{\circ}$, or $240^{\circ}$ as illustrated in the following image:

The second type of symmetries we can look at are axial symmetries along specified axes. There are three axes which we can mirror the equilateral triangle onto itself:

Mirroring the equilateral triangle around each of these axes produces a symmetry:

We will see that these six symmetries form a group. Consider the set $G$ of functions which describes each of these symmetries. One way to describe these symmetries is with functions. For the rotational symmetries, let $\rho_0, \rho_1, \rho_2 : \{ 1, 2, 3 \} \to \{1, 2, 3 \}$ be the rotational symmetry by $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$ respectively.

For each of these functions $\rho_{i} : \{1, 2, 3 \} \to \{1, 2, 3 \}$ for $i =0,1,2$ we will use Cayley's two-line notation described on the Permutations of Elements in a Set as Functions page for which we have that $\rho_i = \begin{pmatrix} 1 & 2 & 3\\ \rho_i(1) & \rho_i(2) & \rho_i(3) \end{pmatrix}$ where the first row denotes the elements in $f$ and the entries in the second row are obtained as the images under each $\rho_i$. Then we have that:

(1)

\begin{align} \quad \rho_0 = \begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix} \quad , \quad \rho_1 = \begin{pmatrix} 1 & 2 & 3\\ 3 & 1 & 2 \end{pmatrix} \quad , \quad \rho_2 = \begin{pmatrix} 1 & 2 & 3\\ 2 & 3 & 1 \end{pmatrix} \end{align}

We can similarly define the axial symmetries $\mu_1, \mu_2, \mu_3 : \{1, 2, 3\} \to \{1, 2, 3 \}$ respectively as:

(2)

\begin{align} \quad \mu_1 = \begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix} \quad , \quad \mu_2 = \begin{pmatrix} 1 & 2 & 3\\ 3 & 2 & 1 \end{pmatrix} \quad , \quad \mu_3 = \begin{pmatrix} 1 & 2 & 3\\ 2 & 1 & 3 \end{pmatrix} \end{align}

Now let $G = \{ \rho_0, \rho_1, \rho_2, \mu_1, \mu_2, \mu_3 \}$, and let $\circ : G \to G$ be defined for each $f_1, f_2 \in G$ as the composition of $f_1$ and $f_2$, that is:

(3)

\begin{align} \quad (f_1 \circ f_2)(x) = (f_2 \circ f_1)(x) \end{align}

Then $(G, \circ)$ is actually a group. To show this, we note that the composition of any two functions in $G$ will indeed be a function that is a symmetry of the equilateral triangle, so $G$ is closed under $\circ$. Furthermore, we have already seen that the composition of three functions is associative, that is, for all $f_1, f_2, f_3 \in G$:

(4)

\begin{align} \quad f_1 \circ (f_2 \circ f_2) = (f_1 \circ f_2) \circ f_3 \end{align}

The identity element in $G$ is $\rho_0$, i.e, the rotation of $0^{\circ}$. Furthermore, each element in $G$ has an inverse in $G$. The pairs of inverses are $(\rho_0, \rho_0)$, $(\rho_1, \rho_2)$, $(\rho_2, \rho_1)$, $(\mu_1, \mu_1)$, $(\mu_2, \mu_2)$, and $(\mu_3, \mu_3)$ as you should verify by what the functions in $G$ mean geometrically.

Therefore $(G, \circ)$ is a group.