Extending a group action to a quotient group If I have a cyclic group $G = (a)$ acting on an abelian group $A$, I need to define a natural action of $G$ on the quotient space $A/B$, where $B$ is a normal subgroup of $A$ with the property that whenever $g\in G$ and $b\in B$, I have $g.b\in B$.
The only natural map that comes to mind is this:
Given a coset $x + B\in A/B$, I want to define $g.(x + B) = g.x + B$.
But I'm having difficulty showing that this action is well-defined.  My problem is that the map $x\mapsto g.x$ is not linear in general.
If $x + B = y + B$, then $y - x\in B$ and so $g.(y-x)\in B$ by assumption.  I want to use this to show that $g.y - g.x\in B$, which would confirm that $g.x + B = g.y + B$.  But since $g.(y - x)\neq g.y - g.x$ I cannot make this jump.
Is this even the action that will work? or am I going down the wrong road?
 A: 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 $g(a) + B \neq g(a^4) + B$.)
