I'm doing Aluffi's Chapter 0, and Exercise 9.13 asks:

Prove that for all subgroups $H$ of a group $G$ and for all $g\in G$, $G/H$ and $G/(gHg^{-1})$ (endowed with the action of G by left-multiplication) are isomorphic in G-Set.

My plan is to first find a bijection between $G/H$ and $G/(gHg^{-1})$, then prove that the map commutes with the action.

The first map I thought of was just $\varphi:G/H \longrightarrow G/(gHg^{-1})$ defined by $aH \mapsto a(gHg^{-1})$, but the problem is that this is not well defined, since we can have $aH = a'H$ but $a(gHg^{-1}) \neq a'(gHg^{-1})$.

What would be well-defined is $aH \mapsto g(aH)g^{-1}$. This works as a bijection between $G/H$ and $G/(gHg^{-1})$, but the problem is that is does not commute with the action of $G$. If the action wasn't strictly left-multiplication, then I could make the map work (by making it conjugate instead), but Aluffi is specifying that the action must be left multiplication.

So now I don't really know what to do. On one hand, my gut feeling tells me that left-multiplication is just not "compatible" with conjugation, and maybe there's no way to make $\varphi$ commute with the action of $G$. On the other hand, Aluffi's exercises are pretty well done, and I might be just missing something?

  • $\begingroup$ Any isomorphism $G/H\to G/gHg^{-1}$ is determined by where it sends $H$ (do you see why?). It must be sent to a coset of $gHg^{-1}$ with stabilizer $H$ (since isomorphisms preserve stabilizers), so it must send $H\mapsto Hg^{-1}$. Then $aH\mapsto aHg^{-1}$. $\endgroup$ – runway44 Apr 6 '20 at 6:15

We want a bijection that is well-defined. If $xH=yH$, then $y^{-1}x\in H$, hence $gy^{-1}xg^{-1}\in gHg^{-1}$, hence $yg^{-1}(gHg^{-1}) = xg^{-1}(gHg^{-1})$.

So, how about mapping the coset $aH$ to the coset $ag^{-1}(gHg^{-1})$?

Why that and not try something like “hence $gy^{-1}g^{-1}gxg^{-1}\in gHg^{-1}$, hence $gxg^{-1}(gHg^{-1}) = gyg^{-1}(gHg^{-1})$; so let’s define the image of $aH$ to be $gag^{-1}(gHg^{-1})$”, which was your second attempt? Because that extra $g$ on the left interferes with the action of $G$, whereas mapping $aH$ to $ag^{—1}(gHg^{-1})$ won’t run into that problem.

Alternatively, you can recognize the map $aH\mapsto ag^{-1}(gHg^{-1})$ as the result of your second map followed by “multiply by $g^{-1}$ on the left”. Essentially, it partially does the conjugation you want to do, but codes it into the bijection instead of into the action.

  • $\begingroup$ Ohhhh. I see I was close. Thank you for your nice explanation. $\endgroup$ – cxx Apr 2 '20 at 22:17
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    $\begingroup$ @cxx: One thing you could try when you run into these issues is to do a small example or two. They can give you the wrong idea just because of the Law of Small Numbers, but they can be suggestive. For instance, trying to do it for $S_3$ and a non-normal subgroup, or $S_4$ and a nonnormal subgroup, might suggest a formula or two for possible bijections. $\endgroup$ – Arturo Magidin Apr 3 '20 at 1:35
  1. Consider $G$ acting on the left cosets of $g^{-1}Hg\leq G$ by the left multiplication
  2. Note that this action is transitive: given $a(g^{-1}Hg)$, $b(g^{-1}Hg)$, we have $a(g^{-1}Hg)=ab^{-1}(b(g^{-1}Hg))$
  3. $H=Stab_G(g(g^{-1}Hg))=Stab_G(Hg)$ since $\forall(h\in H):hHg=Hg$ and $xHg=Hg\Rightarrow xH=H\Rightarrow x\in H$
  4. Apply Proposition 9.9 from the Aluffi's book

P.S. I'm also studying Algebra: Chapter 0 so we can work together

  • $\begingroup$ Hey. It's nice to see another person also working on Aluffi. I don't know how we could work together, though... $\endgroup$ – cxx Apr 6 '20 at 15:51

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