# Group actions as group homomorphisms

The following self-contained post relates to definition 1.2, p. 3, of the book “Groups, graphs and trees: an introduction to the geometry of infinite groups” by John H. Meier.

I have found the interpretation of the definition of group action as a group homomorphism to be verbose. This blog post provides a formalistic exposition of the equivalence between a group action and its induced homomorphism.

Let $G$ be a group and $X$ a set. A group action $\phi$ of $G$ on $X$ is a function $\phi : G \times X \rightarrow X$ that satisfies the properties of

• identity, i.e $(\forall x\in X)\phi(e, x)=x$, where $e$ is the identity element of $G$,
• and compatibility, i.e. $(\forall g, s\in G)(\forall x\in X)\phi(gs, x)=\phi(g, \phi(s, x))$.

For convenience, the shortands $ex=x$ and $(gs)x=g(sx)$ are used in place of $\phi(e, x)=x$ and $\phi(gs, x)=\phi(g, \phi(s, x))$, respectively.

Let $\mbox{Sym}(X)$ denote the symmetric group of $X$, that is the group of bijective functions from $X$ to $X$. Consider the family of bijective functions $\left\{f_g:g\in G\right\}\subseteq\mbox{Sym}(X)$, where $f_g:X\rightarrow X$ and $f_g(x)=\phi(g, x)=gx$.

It is now possible to define the function $h:G\rightarrow \mbox{Sym}(X)$ as $h(g)=f_g$. In words, $h$ maps an element $g\in G$ to the bijective function $f_g:X\rightarrow X$ defined by $f_g(x)=\phi(g, x)=gx$.

The common representation of a group action as a homomorphism relies on the following equivalence; a function $\phi : G \times X \rightarrow X$ is a group action if and only if the function $h:G\rightarrow \mbox{Sym}(X),~h(g)=f_g$, with $f_g(x)=\phi(g, x)=gx$, is a group homomorphism. The main point of this blog post has been to formalize this statement. For the sake of completeness, its proof will be provided, albeit being simple.

Firstly assume that $\phi$ is a group action. It will be shown that $h$ is a homomorphism. By definition, $h(gs)=f_{gs}$. Moreover, $h(g)h(s)=f_gf_s$, where the product $f_gf_s$ denotes function composition, the operation of the symmetric group $\mbox{Sym}(X)$. Since $\phi$ is a group action, it follows that $(\forall x\in X) f_{gs}(x)=(gs)x=g(sx)=(f_gf_s)(x)$, therefore $h(gs)=h(g)h(s)$, so $h$ is a group homomorphism.

Conversely, assume that $h$ is a homomorphism. It will be shown that $\phi$ is a group action.

The identity $e\in G$ belongs to the kernel of $h$; $h(e)=h(ee)$, whence $h(e)=h(e)h(e)$, so $h(e)=\varepsilon$, with $\varepsilon(x)=x$ being the identity function in $\mbox{Sym}(X)$. Furthermore, $h(e)=f_e$, so $f_e=\varepsilon$, which means that $(\forall x\in X)ex=f_e(x)=\varepsilon(x)=x$. The property of identity is thus satisfied for $\phi$.

Since $h$ is a homomorphism, it follows that $h(gs)=h(g)h(s)$, so $f_{gs}=f_g f_s$. Finally, $(\forall x\in X)(gs)x=f_{gs}(x)=(f_gf_s)(x)=g(sx)$, hence the compatibility property of $\phi$ is also satisfied, and $\phi$ is a group action.