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We are given that for a group $G$ and set $S$ denoting the set of subgroups of $G$:

$\rho(g,H) =gHg^{-1}$ where $g \in G, H \leqslant G$ defines a left action of G on S.

We are then asked to determine $Orb(H)$ and $Stab(H)$ for the following subgroups of $G=S_4$:

$H=V_4$, $H=Sym\{1,2,3\}$ and $H=\langle (1234) \rangle$.

Any help would be appreciated. For $H=V_4$, I thought $Orb(V_4)= V_4$ and $Stab(V_4)=S_4$ but this clearly doesn't agree with the Orbit-Stabiliser theorem so not too sure where I've gone wrong.

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  • $\begingroup$ Isn't the orbit-stabilizer theorem for one element? $\endgroup$ – J. W. Tanner May 15 '20 at 17:07
  • $\begingroup$ @J.W.Tanner an element of the set S. here S is the set of subgroups and so V4 is an element of the set. $\endgroup$ – DietCola01 May 15 '20 at 17:09
  • $\begingroup$ Ya, sorry, I had missed that, it is clearly stated. I have deleted my previous comment. $\endgroup$ – user750041 May 15 '20 at 17:16
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For $H=V_4$, I thought $\mathrm{Orb}(V_4)=\{V_4\}$ and $\mathrm{Stab}(V_4)=S_4$ but this clearly doesn't agree with the Orbit-Stabiliser theorem so not too sure where I've gone wrong.

You're wrong about being wrong. There is no disagreement with the orbit-stabizer theorem. Since the stabilizer is the whole group, the size of the quotient $G/\mathrm{Stab}$ is $1$, which is the size of the orbit $\{V_4\}$ (notice we must put $V_4$ in parentheses, since it is a single element of the orbit).

Any ideas for $H=S_3$ and $H=\langle (1234)\rangle$?

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  • $\begingroup$ Oh I see thank you! For $H=S_3$ I was thinking it may have something to do with $S_4/V_4 \cong S_3$ (this was shown earlier on) but not too sure from there and the last one again i'm not too sure $\endgroup$ – DietCola01 May 15 '20 at 18:46
  • $\begingroup$ @DietCola01 That's a good fact to know, but not relevant at the moment. Since you're working with small groups, you can get your hands dirty by working out concrete examples. For instance if $g=(1234)$ and $H=\mathrm{Sym}(\{1,2,3\})$ then what are the elements of $gHg^{-1}$; can you describe $gHg^{-1}$ any way other than listing its elements? $\endgroup$ – runway44 May 15 '20 at 18:55
  • $\begingroup$ this was very helpful thank you! $\endgroup$ – DietCola01 May 15 '20 at 19:02

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