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Please could someone check my proof that $\varphi : G(x) \to G/G_x$ is injective?

The notation is the following: $G$ is a group acting on a set, $G_x = \{g \in G\mid gx = x \}$ and $G(x) = \{gx \mid g \in G\}$. Define

$$ \varphi: G(x) \to G/G_x, \hspace{0.5cm} gx\mapsto gG_x$$

My proof:

Assume that $\varphi (gx) = \varphi (hx)$. Then $gG_x = hG_x$. That is, $$\{ gg' \mid g' x = x \} = \{hg' \mid g' x = x\}$$

In particular, there exist $g'$ and $g''$ with $g' x = x$ and $g'' x = x$ and $gg' = hg'$. Then $gg'(g'')^{-1} = h$ and $g(g'')^{-1} x= x$. Then $$ gx = gg' (g'')^{-1} x = hx$$ hence $gx = hx$ hence $\varphi$ is injective.

Jendrik Stelzner
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user174981
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    You need to show that $\varphi$ is well defined. That is, if $gx=hx$, you need to prove that $\varphi(gx)=\varphi(hx)$. One always has to do this when, instead of defining a function with a "generic" argument, one assumes a particular form of argument (here, instead of taking any $y$ in $G(x)$, you decompose it into $gx$ but the choice of $g$ is not canonical, maybe multiple $g$ would have worked). – zarathustra Jan 17 '15 at 08:00
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    The proof of injectivity is fine. Though you could do it more neatly by deducing from $gG_x=hG_x$ that $h^{-1}g\in G_x$. – caffeinemachine Jan 17 '15 at 08:00
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    @zarathustra In most cases, the way $\varphi$ has been defined does require well-definedness to be explicitly shown. But since for each element $a$ of $G_x$ there is a unique $g\in G$ such that $gx=a$, here the ell-definedness, in my opinion, may be omitted. – caffeinemachine Jan 17 '15 at 08:04
  • @coffeinemachine "But since for each element $a$ of $Gx$ there is a unique $g\in G$ such that $gx=a$": this is only true if the stabilizer of $x$ is trivial! – zarathustra Jan 17 '15 at 08:08
  • @zarathustra Ok, I now also showed that it's well-defined. It was not difficult. – user174981 Jan 17 '15 at 10:00

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You can also proceed differently by starting with $G → Gx,~g ↦ gx$ which is (by definition) surjective and noting that $gx = hx ⇔ h^{-1}gx = x ⇔ h^{-1}g ∈ G_x$, hence the map factors injectively through $G/G_x$ maintaining surjectivity.

k.stm
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  • What does "maintaining surjectivity" mean? – user174981 Jan 17 '15 at 09:54
  • @user174981 It means that the map $G/G_x → Gx,~[g] ↦ gx$ coming from the surjective map $G → Gx,~g ↦ gx$ is still surjective. – k.stm Jan 17 '15 at 10:30
  • But... how did you prove that the map stays surjective after factoring? – user174981 Jan 17 '15 at 10:49
  • @user174981 Let $y ∈ Gx$. From the surjectivity of $G → Gx$ or the definition of $Gx$, there is a $g' ∈ G$ such that $g'x = y$. So $[g'] ∈ G/G_x$ maps to $y$ as well under $G/G_x → Gx,~[g] ↦ gx$, proving the surjectivity of the map. – k.stm Jan 17 '15 at 10:53
  • https://math.stackexchange.com/questions/21932/what-does-it-mean-to-say-a-map-factors-through-a-set – tryst with freedom Mar 11 '22 at 00:25