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 be a group and a set. A group action of on is a function that satisfies the properties of

- identity, i.e , where is the identity element of ,
- and compatibility, i.e. .

For convenience, the shortands and are used in place of and , respectively.

Let denote the symmetric group of , that is the group of bijective functions from to . Consider the family of bijective functions , where and .

It is now possible to define the function as . In words, maps an element to the bijective function defined by .

The common representation of a group action as a homomorphism relies on the following equivalence; a function is a group action if and only if the function , with , 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 is a group action. It will be shown that is a homomorphism. By definition, . Moreover, , where the product denotes function composition, the operation of the symmetric group . Since is a group action, it follows that , therefore , so is a group homomorphism.

Conversely, assume that is a homomorphism. It will be shown that is a group action.

The identity belongs to the kernel of ; , whence , so , with being the identity function in . Furthermore, , so , which means that . The property of identity is thus satisfied for .

Since is a homomorphism, it follows that , so . Finally, , hence the compatibility property of is also satisfied, and is a group action.

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