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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?

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It seems like you have proved that you get the desired action if and only if the action of $G$ on $A$ is linear. So you either have to assume that the action of $G$ is linear (what I would recommend) or come up with a counterexample when the action is not linear. But I really think the point of your observation is that you also need to assume the action is linear (in fact, most people would include that in the definition of a group $G$ acting on an abelian group $A$; if they didn't want a linear action they'd just talk about $G$ acting on the set $A$). –  Michael Joyce Sep 10 '12 at 20:33
    
Actually, this is a problem from Serge Lang's Algebra (3rd) (Chapter 1, problem 45), and the quote from the problem is "Assume that $G$ acts on an abelian group $A$ (just as sets)..." –  Kyle Schlitt Sep 10 '12 at 20:41
    
IT actually asks me to define an operation of $G$ on $A/B$, but I interpreted this as action. –  Kyle Schlitt Sep 10 '12 at 20:43
    
Perhaps the most natural way of thinking about this is that if $G$ acts on $A$ then there is a homomorphism $G\rightarrow \operatorname{Aut}(A)$. Then find the subgroup of $\operatorname{Aut}(A)$ which $G$ surjects onto, $H$ say, and prove that $H\rightarrow A/B$ . In my eyes, this is neater than fooling around with actions... –  user1729 Sep 11 '12 at 11:04
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1 Answer 1

up vote 4 down vote accepted

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 $g(a) + B \neq g(a^4) + B$.)

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Thanks for editing, M Turgeon. For some reason, the LaTeX isn't showing up on my screen - is this a known bug? –  Billy Sep 10 '12 at 21:05
    
Thanks for this. Your answer, and the above comment from Michael Joyce suggests that I should assume that $g.(x + y) = g.x + g.y$ for all $x,y\in A$, $g\in G$. This makes my map work, and solves my problem. Thanks for taking the time to come up with the counter-example! –  Kyle Schlitt Sep 10 '12 at 21:19
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