Group Action and a Cartesian Product

[chaotixmonjuish]

Suppose a group G and it acts on a set X and a set Y.

(a) A simple group action on the cartesian product would be defined as such:

G x (X x Y) --> (X x Y)

to prove this is a group action could I just do this:

Suppose a g₁ and g₂ in G. g₁*(g₂*(x,y))=g₁*g₂(x). This is obvious. Basically is the proof extremely easy. I just grabbed this example out of a book and was wondering if I am close.

 

[wofsy]

Suppose a group G and it acts on a set X and a set Y.

(a) A simple group action on the cartesian product would be defined as such:

G x (X x Y) --> (X x Y)

to prove this is a group action could I just do this:

Suppose a g₁ and g₂ in G. g₁*(g₂*(x,y))=g₁*g₂(x). This is obvious. Basically is the proof extremely easy. I just grabbed this example out of a book and was wondering if I am close.

Not sure what your question is but you seem to have a typo in your last equation.

g.(x,y) = (g.x,g.y) works on the direct product. Verify this directly.

 

[chaotixmonjuish]

I guess the question is asking for me to define and prove a very general and simple action on that cartesian product.

 

[bober]

Yeah, basically you just need to show that g1*(g2*(x))=(g1*g2)(x) and g1*(g2*(y))=(g1*g2)(y), which is really what the definition states. I think.

 

[chaotixmonjuish]

For the sake of group theory, when g operates on y in Y, would it need to be using a different operator such as '$'