I refer to 'Analogy of ideals with Normal subgroups in groups' which was a very enlightening question for me. When I was young I was too avid on abstract algebra and I did too many courses at the same time. Which of course resulted in a great deal of lack of understanding. I came to love Emmy Noether and her three theorems of isomorphisms for groups and did well on that higher level, but always felt insecure about the details of the cosets. Finally, thanks to the intuition of the questioner, I think I have managed to conquer and secure the subject.

The group $G$ is acting on the normal subgroup $N$ as $$ \phi:G\times N\rightarrow N\qquad (n'=g^{-1}ng). $$ The ring $R$ is acting on the ideal $A$ as $$ \psi:R\times A\rightarrow A\qquad (a'=ra). $$ $\phi_g(n)=\phi(g,n)$ makes $\phi_g:N\rightarrow N$ to a homomorphism because

$$ \phi_g(nm)=g^{-1}(nm)g=g^{-1}(ngg^{-1}m)g=(g^{-1}ng)(g^{-1}mg)=\phi_g(n)\phi_g(m) $$ and there is a homomorphism $\theta:G\rightarrow End(N)$, where $g\mapsto \phi_g$, since $$ \phi_g\phi_h(n)=\phi_g(\phi_h(n))=\phi_g(h^{-1}nh)=(g^{-1}h^{-1})n(hg)=(hg)^{-1}n(hg)=\phi_{hg}(n) $$

Is my intuition right when claiming: $Im\, \theta \simeq G/N$? And similar for rings?

Can quote objects be defined in a similar way in some other categories?

  • $\begingroup$ What is a "quote object"? Do you mean "quotient object"? Maybe you want to think about the quotient of a set by an equivalence relation: that generalizes quite well, and you still have the three isomorphism theorems. Both normal subgroups and ideals are special cases: they are fancy ways of expressing an equivalence relation. In even greater generality, coequalizers. $\endgroup$ – Hurkyl Sep 2 '14 at 14:56
  • $\begingroup$ Yes I meant quotient objects but not only equivalence classes. But it didn't work. I will check the coequalizers. $\endgroup$ – Lehs Sep 2 '14 at 16:06
  • $\begingroup$ @Hurkyl: which are the 3 isomorpisms for equivalence classes? Maybe I should look at that to. $\endgroup$ – Lehs Sep 2 '14 at 18:11
  • $\begingroup$ wikipedia talks about the isomorphism theorem for equivalence classes. (actually for congruences; but for sets, all equivalence relations are congruences) $\endgroup$ – Hurkyl Sep 2 '14 at 22:00
  • $\begingroup$ @Hurkyl: Yes, eq-relations seems to be the right approach. Thank you! $\endgroup$ – Lehs Sep 3 '14 at 5:53

No. This doesn't work.

For example: Let $G=\mathbb{Z}$ (under +) and $N = 2\mathbb{Z}$ (even integers). Then $G/N$ is the group of order 2 (i.e. integers mod 2). However, $\phi_g(n)=-g+n+g=n$ so $\phi_g$ is the identity map for each $g \in G$. Thus $\mbox{Im}(\theta)$ would be the trivial group!

What is true is that the kernel of $\theta$ is $g \in G$ such that $\phi_g$ is the identity map. This is precisely the set of $g \in G$ that commute with all $N$ -- that is -- the centralizer of $N$ in $G$: $\mbox{ker}(\theta) = C_G(N) = \{ g \in G \;|\; gn=ng \mbox{ for all } n \in N\}$. So the first isomorphism theorem states that $G/C_G(N) \cong \mbox{Im}(\theta)$.

Does this generalize? Yes and no. You can state something similar for rings if you make certain assumptions. There isn't a direct generalization though...and certainly not for a general category. :(

Edit: For rings...

The map $\phi_r(a)=ra$. Then $\phi_r$ is a homomorphism of $R$ (thought of as a group under addition -- $\phi_r$ is not a ring homomorphism in general). We can see that $\phi_{r+s}=\phi_r+\phi_s$ and $\phi_{rs}=\phi_r \circ \phi_s$. So $\theta$ gives us a ring homomorphism from $R$ to $\mbox{End}(R)$ where $\mbox{End}(R)$ is the collection of endmorphisms of the abelian group $(R,+)$. Of course $\mbox{End}(R)$ becomes a ring when given the operations of adding homomorphisms pointwise and composing homomorphisms.

In this case, the kernel of $\theta$ is all $r \in R$ such that $\phi_r$ is the zero map on $A$. This is called the annihilator of $A$ and denoted $\mbox{Ann}_R(A)$. The first isomorphism theorem gives us that $R/\mbox{Ann}_R(N) \cong \mbox{Im}(\theta)$.

As you can see, this is quite a bit different from your group example.

  • $\begingroup$ You are right of course. Thanks! $\endgroup$ – Lehs Sep 2 '14 at 16:07
  • $\begingroup$ Thank you once again. I was not thinking of morphisms of rings but of modules. But it doesn't work, my intuition was wrong and I'm lucky to have someone else to do the thinking. :) $\endgroup$ – Lehs Sep 2 '14 at 18:09

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