Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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)..." – roo 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. – roo 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
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 $g(a) + B \neq g(a^4) + B$.)

share|cite|improve this answer
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! – roo Sep 10 '12 at 21:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.