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.

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> Prove that: $S\_n$ acts on {1, 2, . . . , n} via application of each map. Where 
>  $S\_n$ is the symetric group.
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Also:

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