How can I prove that a group acts on these sets:

Question

Suppose that G is a group and X is a set. An action of G on X is a map G× X → X where $(g, x) → g \circ x$ such that both

(a) $1_G \circ x = x$ for each x ∈ X and (b) $g \circ (h \circ x) = (gh) \circ x$ for all g, h ∈ G and x ∈ X.

Prove that: $S_n$ acts on {1, 2, . . . , n} via application of each map. Where $S_n$ is the symetric group.

Also:

Prove that: A group G acts on itself (as X) by left multiplication: $g \circ x = gx$

For this one I guess I solved it alright by doing: $g \circ (h \circ x)=ghx$ and because the group operation is associative it follows that $g \circ (h \circ x)=(gh)\circ x$ This would prove (b), I would want an opinion if it's right.

Prove: A group G does not act on itself (as X) by right multiplication: $g \circ x = xg$

For this one I thought that $g \circ (h \circ x)=xhg$ which is not possible because xhg can't be rearanged so that is equal to $(gh)\circ x$

But again it is not concrete for me what I am saying so it might be wrong.

I know that I must prove both criteria from above, but I stuck.

Answer 1

For the first part, id(x)=x and for permutations, we know that $g(h(x))=gh(x)$ so, its true.

Your proof for the second part is correct.

For the third part, consider $g\circ (h\circ x)=xhg\neq gh\circ x=xgh$ in general unless gh=hg

Answer 2

For the last two questions, I don't think they are correct. For example, if the group multiplication is defined by $ g \circ x = g^{-1}x $, then $ g \circ h\circ x = g \circ (h^{-1}x) = g^{-1}h^{-1}x = (hg)^{-1}x = (hg)\circ x $, which is indeed a right action.