# Group Action and a Cartesian Product

## Main Question or Discussion Point

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.

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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.

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

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.

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