# Isomorphism of a quotient group using the stabilizer

It seems that $G_s$ need not be normal to have an isomorphism ($G_s$ and $O_s$the are stabilizer and orbit of a group action):

$G/G_s \cong O_s$

First, is this accurate? Then I was wondering how this reconciles with the First Isomorphism Theorem where the divisor, as the kernel of the map, is normal.

In the case of the group action, an element of $G_s$ seems to be an element of the kernel, since $g(s) = 1s$. But $g$ $G_s$ $g^-1$ is the stabilizer $G_t$, where $g(s) = t$?

I can't imagine that I have not made several mistakes here, so I appreciate your patience as well as setting me straight. Thanks.

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Well, if $G_s$ isn't normal, then $G/G_s$ isn't really a group, since the usual operation on cosets isn't well-defined. –  user5137 Nov 29 '11 at 1:01
Also, $O_s$ is prettymuch never a subgroup of $G$, since orbits partition $G$ (so the identity of $G$ is in exactly one orbit). –  user5137 Nov 29 '11 at 1:09

You cannot get the whole of the First Isomorphism Theorem from the Orbit-Stabilizer Theorem, because the latter ignores any structure that your set $S$ may carry, whereas the First Isomorphism Theorem very much cares about the structure of your set $H$.

But you can do the following: given a homomorphism $f\colon G\to H$, you can make $H$ into a $G$-set by the rule $g\cdot h = f(g)h$.

This defines an action, since $e\cdot h = f(e)h = h$, and $$(xy)\cdot h = f(xy)h = \bigl(f(x)f(y)\bigr)h = f(x)\bigl(f(y)h\bigr) = x\cdot(y\cdot h).$$

Now consider the identity $1\in H$. The orbit of $1$ is precisely the image of $f$; the stabilizer of $1$ is $$G_1 = \{g\in G\mid f(g)1 = 1\} = \{g\in G\mid f(g)=1\} = \mathrm{ker}(f).$$

So the Orbit-Stabilizer Theorem tells you there is a bijection between cosets $G/\mathrm{ker}(f)$ and $f(G)$ given by $g(\mathrm{ker}(f)) \mapsto f(g)$.

However, the Orbit-Stabilizer Theorem does not tell you that this bijection respects the group structures on $G/\mathrm{ker}(f)$ and on $H$. You have to prove this explicitly. So the Orbit-Stabilizer Theorem only gives you the set-theoretic bijection, you still need to prove that it is a group isomorphism.

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This is exactly the kind of thing I meant (and couldn't come up with) -- thank you! –  sibilant Nov 29 '11 at 4:39

Let $G$ be a group and $S$ a set with an action of $G$. I'll assume it's a left action, so that $g(h(s)) = (gh)(s)$. If $O_s \subset S$ is the orbit of $s \in S$ (that is, $O_s = { g(s): g \in G }$), and $G_s \subset G$ is the stabilizer of $s$ (so $G_s = \{g \in G: g(s) = s \}$, a subgroup of $G$), then there's an isomorphisms of sets-with-$G$-action

$$G/G_s \to O_s$$

given by $gG_s \mapsto g(s)$. You can check that this map is well-defined: if $gG_s = hG_s$, then $g(s) = h(s)$.

As you noted, $G_s$ is not a normal subgroup of $G$ (yes, $gG_s g^{-1} = G_{g(s)}$), so what I mean by $G/G_s$ here is the space of left cosets $\{g G_s: g \in G\}$ of $G_s$ in $G$.

This coset space has a left action of $G$ by translation (that is, $g(hG_s) = gh G_s$), as does the set $O_s$ (the action is inherited from the action on $S$) and the map above is $G$-equivariant (starting with a left coset, you can act by $g$ and then apply the map or apply the map and then act by $g$, and get the same element of $O_s$ either way).

The First Isomorphism Theorem is a statement about a map of groups $f: G \to H$, not about an action of a group on a set, so a priori is dealing with a different scenario. You can probably find ways to see one of these reducing to the other in special cases, but no particularly illuminating setups come to mind at the moment...

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@sibilantthanks. I've been especially puzzling over what you mention in the last paragraph, primarily based on the visual appearance X/Y isomorphic to Z –  Andrew Nov 29 '11 at 2:28
Well, the similarity in appearance isn't meant to be deceiving. In both cases, $X/Y$ is the space of left cosets, and the isomorphism with $Z$ preserves as much structure as it can. It's just that in the orbit-stabilizer case, $X/Y = G/G_s$ is just a set (or pointed set) with an action of $G$, and in the first isomorphism theorem $X/Y = G/\ker f$ has a group structure (and, if you like, still an action of $G$, as in Arturo Magidin's answer). –  sibilant Nov 29 '11 at 4:47

It was already mentioned in the other answers that $G/G_s$ is not meant to be a group but only a $G$-set, and that $G/G_s\cong Gs$ is an isomorphism of $G$-sets (I write $Gs$ for the orbit). I just want to stress the analogy with modules over a (unitary) ring $R$: given a module $M$ and an $m\in M$, the map $r\mapsto rm$ induces an isomorphism $R/I\cong Rm$ of $R$-modules via the isomorphism theorem for module, where $I=\{ r\in R \mid rm=0 \}$. Again, $I$ is a left ideal but may not be an ideal and $R/I$ is not meant to be a ring.

This analogy also suggest that one can obtain the isomorphism $G/G_s\cong Gs$ also via an isomorphism theorem, namely the one for sets:

If $f: X\rightarrow Y$ is a map of sets, one has the kernel relation

$\kappa(f) = \{ (x,x')\in X\times X \mid f(x)=f(x') \} \subseteq X\times X$

and there is a unique map $\bar{f}: X/{\kappa(f)} \rightarrow Y$ with $\bar{f}([x]) = f(x)$ where $X/{\kappa(f)}$ is the set of equivalence classes and $[x]$ is the equivalence class of $x$. This map is injective, and if $f$ is surjective then $\bar{f}$ is also surjective, hence an isomorphism (of sets).

Now, if $X$ and $Y$ are $G$-sets and $f$ is a morphism of $G$-sets, then $\kappa(f)$ is invariant under the action of $G$ (because $f(x)=f(x')$ gives $f(gx) = g(f(x)) = g(f(x')) = f(g(x'))$, the set $X/{\kappa(f)}$ can be given an $G$-action with $g[x] = [gx]$, and the map $\bar{f}$ is a morphism of $G$-sets. This is your situation here.

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