My understanding of the phrase "$G$ acts on $A$" is something along the lines of:
- elements of $G$ are behaving like functions $A \to A$ (edit: and the identity element behaves like the identity function),
- composition of these functions behaves like multiplication in $G$,
- each function respects the structure on $A$.
That is, if $A$ is a set, then the third stipulation is meaningless, but if $A$ is a group, then I expect these functions to be group homomorphisms. Likewise if $A$ is a topological space, I expect these functions to be continuous maps. A shorter way of saying all this: an "action" of $G$ on $A$ is simply a group homomorphism $G \to \mathrm{Aut}(A)$, whatever an 'automorphism' of $A$ means.
(If $G$ is acting on $A$ purely as a set, I don't think this can be true. Let $G = \{1, g\}$, $A = \{1, a, \ldots, a^5\}$, $B =\{ 1, a^3\}$, all cyclic groups. Now suppose $g$ acts on $A$ by swapping $a$ and $a^2$, but leaving the other four points fixed. Then $a + B = a^4 + B$, but $g(a) + B \neq g(a^4) + B$.)
a)The action that is transitive and faithful
Your Answer: Group G under addition acting on a set of integers Z
It's not an answer because you did not say how it is acting (see also Leon Aragones'comment). By the way, $\mathbb{Z}$ is always acting in a natural way on $G$ when $G$ is abelian (but the action might not be transitive).
My Hint (trivial) : The trivial group $G$ acts on $\{1\}$ trivially, this should do the job why ?
My Hint (a little less trivial) : Take $G$ a group acting on itself in some particular way...
b)The action that is transitive but is NOT faithful
Your Answer: Group G of 60 degree rotations acting on a set of Vertices of Dihedral group $D_3$ since all 3 rotations fix everything.
"set of Vertices of Dihedral group $D_3$" do you mean the set of vertices of the equilateral triangle $T$ for which $D_3=Isom(T)$? In that case there are 2 rotations which does not fix everything.
My Hint (trivial) : You should go more simple, take $X:=\{1\}$ (i.e. a set with one element) then a group $G$ always acts on $X$ (the action is unique why?). Under which conditions on $G$ is the action transitive? faithful?
My Hint (a little less trivial) : Take $G$ be a group with at least one proper subgroup $H$ which is not trivial then $G$ acts naturally on $G/H$ (left $H$-cosets). This should do the job.
c)The action that is NOT transitive but IS faithful
Can I simply say it's just a group of symmetries of a single point? Since it only has 1 element it can't be transitive, right?
It seems about right but you do need to give the whole set up.
My Hint (trivial) : Take $G$ to be the trivial group and $X$ be any set. Then there is only one action of $G$ on $X$. Under which condition on $X$ is the action transitive? faithful?
My Hint (a little less trivial) : Take $G:=GL_2(\mathbb{R})$, I claim that there is a natural action on $\mathbb{R}^2$, this one do the job.
d) The action that is NOT transitive and NOT faithful
Something that is non abelian? I don't really know.
Go simple.
My Hint (trivial) : take $G$ a group and $X$ a set. Define the action of $G$ on $X$ by $g.x:=x$. Under which conditions on $G$ and $X$ is the action transitive? faithful?
My Hint (a little less trivial) : Take $G$ be a group with at least one proper subgroup $H$ which is not trivial then $G$ acts naturally on $G/H$ (left $H$-cosets). Then $G$ also acts diagonally on $G/H\times G/H$. This should do the job.
e) The action with 2 orbits
A line with vertices 1 and 2. - The group of symmetries acting on a set of vertices.
In that case the action has only one orbit, hasn't it?
My Hint (trivial) : Trivial group and a set with two elements.
My Hint (a little less trivial) : Think about a square and an axial symmetry.
f) The action with 3 orbits
I have a triangle in my mind, but rotational symmetries of a triangle are stabilizers aren't they? Same goes for part (e).
My Hint (trivial) : Trivial group and a set with three elements.
My Hint (a little less trivial): Think about an hexagon and an axial symmetry.
Best Answer
Yes. The induced action makes the coset of $g$ map the orbit of $x$ to the orbit of $gx$.
That is: if $[x] = \{nx\mid n\in N\}$, then we let $(gN)[x] = [gx]$.
To verify that this is well defined, first note that if $[x]=[y]$ and $g\in G$, then there exists $n\in N$ such that $x=ny$. Then $gx = gny$. But since $N$ is normal, $gN=Ng$, so there exists $n’\in N$ such that $gn=n’g$. Therefore, $gny = n’gy$. Therefore, $[gx]=[gy]$.
And second, if $gN=hN=Nh$, then there exists $n\in N$ such that $g=nh$. Therefore, $gx = nhx$, so $[gx]=[hx]$. Thus, $(gN)[x] = [gx]=[hx] = (hN)[x]$ as required.
Finally, this is indeed an action: $eN[x] = [ex] = [x]$ and $((gN)(hN))[x] = (gh)N[x] = [(gh)x] = [g(hx)] = gN[hx] = (gN)(hN[x])$.
Note that you do not need $N$ to be abelian, just normal.