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Question

I am trying to understand the difference between a group action and left multiplication of a group. If a group action is defined as $$F: G\ \times X\ \longrightarrow X$$

then I interpret this to mean that G is a group and X is a set. I have been working through a paper explaining this definition in the following terms:

"Note that a group action is not the same thing as a binary structure. In a binary structure, we combine two elements of X to get a third element of X (we combine two apples and get an apple). In a group action, we combine an element of G with an element of X to get an element of X (we combine an apple and an orange and get another orange)."

However, later in the paper, it is stated that:

"If G is a group, then G acts on itself by left multiplication: g · x = gx. The axioms of a group action just become the fact that multiplication in G is associative (g1(g2x) = (g1g2)x) and the de?nition of the identity (1x = x for all x ∈ G). More generally, if H ≤ G is a subgroup, not necessarily normal, then G acts on the set of left cosets G/H via: g · (xH) = (gx)H. The argument that this is indeed an action is similar to the case of left multiplication."

My confusion is that if G and H are "apples and oranges", and if a group action is "not the same thing as a binary structure", then how can "a group acting on itself by left multiplication" be a group action? Would that not render a definition of a group action as $$F: G\ \times G\ \longrightarrow G$$ ?

I am aware of an axiomatic way of defining a group action, namely by associativity and identity, such that

for all $g1,g2\in G $ and $ x \in X, g1\cdot (g2\cdot x) = (g1g2) \cdot x $

and

for all $x\in X,$ $1\cdot x=x$

Am I on the right track by assuming that any apples or oranges satisfying these axioms will define a group action, even if the apples are oranges? If anyone can give a clarification using an example of a group acting on cosets, that would be very generous, as that is the specific case I am tying to understand.

Thanks for any help, and my apologies for any noob mistakes in notation or terms.

Answer 1

If $G$ is a group then it is a set together with a binary operation. So in particular, we have a set here. So $G$ can act on the set $X=G$, just as it can act on any other set.

If it confuses you, denote $X=G$. We can define an action $F:G\times X\to X$ by $(g,x)\to gx$ where $gx$ is the product in $G$. It's very easy to check that this is indeed a group action. So in this specific example $F$ as a function indeed coincides with the binary operation in $G$. But that's not a contradiction.