# Understanding the concept of group action via an example

The following self-contained post clarifies the definition of group action (definition 1.2, p. 3, in the book “Groups, graphs and trees: an introduction to the geometry of infinite groups” by John H. Meier) by adapting an example from chapter 7, p. 72-74, of the book “A book of abstract algebra” by Charles C. Pinter.

A rigorous definition of group action and of its group homomorphism representation was provided in a previous post. This post presents the group action on the symmetries of the square to exemplify the concept.

Consider a square with numbered vertices, as shown in the figure below. The set of vertices of the square is $X=\left\{1,2,3,4\right\}$.

A symmetry of the square can be informally thought of as a way of moving the square so that it coincides with its former position. Every such move is fully described by its effect on the vertices, in the sense that every new vertex position coincides with a distinct former vertex position.

There are exactly 8 symmetric moves of the square, each of which can be described by a permutation of the square’s vertices. Here is the list of the 8 symmetries of the square:

• The identity $R_0=\left(\begin{matrix}1 & 2 & 3 & 4\\ 1 & 2 & 3 & 4 \end{matrix}\right)$, which does not move the square.
• Clockwise rotation of the square about its centre $P$ by an angle of $90^{\circ}$$R_1=\left(\begin{matrix}1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1\end{matrix}\right)$.
• Clockwise rotation of the square about its centre $P$ by an angle of $180^{\circ}$$R_2=\left(\begin{matrix}1 & 2 & 3 & 4\\ 3 & 4 & 1 & 2\end{matrix}\right)$.
• Clockwise rotation of the square about its centre $P$ by an angle of $270^{\circ}$$R_3=\left(\begin{matrix}1 & 2 & 3 & 4\\ 4 & 1 & 2 & 3\end{matrix}\right)$.
• Flip of the square about its diagonal $A$ (see figure below): $R_4=\left(\begin{matrix}1 & 2 & 3 & 4\\ 1 & 4 & 3 & 2\end{matrix}\right)$.
• Flip of the square about its diagonal $C$$R_5=\left(\begin{matrix}1 & 2 & 3 & 4\\ 3 & 2 & 1 & 4\end{matrix}\right)$.
• Flip of the square about its vertical axis $B$$R_6=\left(\begin{matrix}1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3\end{matrix}\right)$.
• Flip of the square about its horizontal axis $D$$R_7=\left(\begin{matrix}1 & 2 & 3 & 4\\ 4 & 3 & 2 & 1\end{matrix}\right)$.

The set of all 8 symmetries of the square is denoted by $D_4=\left\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\right\}$. Define the operation $\circ : D_4\times D_4\rightarrow D_4$ to be the function composition $R_i\circ R_j(x)=R_i(R_j(x))$ for $x\in X\left\{1,2,3,4\right\}$. For example,

$R_1\circ R_6$ is the result of first flipping the square about its vertical axis $B$ and then rotating it clockwise by $90^{\circ}$:

$R_1\circ R_6 =\left(\begin{matrix}1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1\end{matrix}\right)\circ\left(\begin{matrix}1 & 2 & 3 & 4\\ 2 & 1 & 4 & 3\end{matrix}\right)=\left(\begin{matrix}1 & 2 & 3 & 4\\ 3 & 2 & 1 & 4\end{matrix}\right)=R_5$.

$R_1\circ R_6 =R_5$ means that first flipping the square about its vertical axis $B$ and then rotating it clockwise by $90^{\circ}$ is the same as flipping the square about its diagonal $C$.

The set of symmetric permutations $D_4$ of the square along with the operation of permutation composition induces the so-called dihedral group $(D_4, \circ)$ of the symmetries of the square. $|D_4|$ denotes the order of group $(D_4, \circ)$, which is the number of elements of $D_4$. Obviously, $|D_4|=8$.

The symmetric group $(\mbox{Sym}(X), \circ)$ of $X=\left\{1,2,3,4\right\}$ is the set of all the permutations of $X$, i.e. the set of all the bijective functions from $X$ to $X$. Since $\mbox{Sym}(X)$ has $4!$ elements, the order of $\mbox{Sym}(X)$ is $|\mbox{Sym}(X)|=24$.

Note that $D_4\subseteq \mbox{Sym}(X)$ and that $(D_4, \circ)$ is a subgroup of $(\mbox{Sym}(X), \circ)$.

Any function $\phi:D_4\times X\rightarrow X$ is a group action as long as it satisfies

• $\phi(R_0, x)=x$ for all $x \in X=\left\{1,2,3,4\right\}$ (identity property) and
• $\phi(R_i\circ R_j, x)=\phi(R_i, \phi(R_j, x))$ for all $i, j=0,1,2,3,4,5,6,7$ and for all $x \in X=\left\{1,2,3,4\right\}$ (compatibility property).

One way of picking a specific group action $\phi$ relies on defining the type of associated group homomorphism $h:D_4\rightarrow\mbox{Sym}(X)$ in a way that respects the identity and compatibility properties of $\phi:D_4\times X\rightarrow X$.

The simplest possible example would be to choose the group homomorphism $h:D_4\rightarrow\mbox{Sym}(X)$ to be the identity function $h(R_i)=R_i,~i=0,1,2,3,4,5,6,7$, in which case the group action $\phi$ takes the form $\phi(R_i, x)=R_i(x)$.

It is easy to check that the the group action $\phi(R_i, x)=R_i(x)$, which arises by setting the group homomorphism $h$ to be the identity function, satisfies the properties of identity and compatibility:

• $\phi(R_0, x)=R_0(x)=x$,
• $\phi(R_i\circ R_j, x)=R_i\circ R_j(x)=R_i(R_j(x))=R_i(\phi(R_j,x))=\phi (R_i,\phi (R_j, x))$.

It is also easy to see for instance that the group action $\phi(R_i, x)=R_i(x)$ maps $(R_1, 2)\in D_4\times X$ to $3\in X$, since $\phi(R_1, 2)=R_1(2)=3$.

The group action $\phi(R_i, x)=R_i(x)$ is interpreted as the function that maps every symmetric move (permutation) $R_i$ of the square and every square vertex $x$ to the square vertex $R_i(x)$.

The group homomorphism $h(R_i)=R_i$ is the identity, so it is an injective function. As elaborated in this previous post, since $h$ with $h(R_i)=R_i$ is injective, the group action $\phi$ with $\phi(R_i, x)=R_i(x)$ is faithful.