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$G = S_3$ I need to examine the operation of conjugation of $G$ on the group of the subgroups of $G$ (meaning $gH = gHg^{-1}$)

here I already got confused from the definition.

Now I need to : a) find the group of orbits $O$ of this operation.

b) for each orbit $o \in O$ choose a representative $H \in o$ and calculate $\operatorname{Stab}_G(H)$.

c) check the Orbit-stabilizer theorem on this operation.

I'm really confused from the definitions here.

What I tried - If I understand the question correctly - I have H' which is a group that contains 6 subgroups of $S_3$ then i calculate the orbits for each of these groups. I get one orbit for each group. So I guess here the representative (asked in (b) ) is this one orbit that I found. for each of the 6 orbits I get that

$$\operatorname{Stab}_G(O_i) = S_3$$

so in (c) i get that

$|H'| = \sum_1^6 [G:\operatorname{Stab}_G(x_i)] = 1+1+1+1+1+1 = 6$ which is correct but I'm sure i did not understand the question correctly . any help will be very appreciated!

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  • $\begingroup$ It's probably confusing to use $gH$ for the conjugation action of $g$ on subgroup $H$, when its definition involves $gH$ with its usual meaning. $\endgroup$ – pjs36 Sep 5 '15 at 17:33
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First of all I guess what you write as conjunction you mean conjugation (at least this is the name I know). Secondly your $H'$ is the set of subgroups of $S_3$. ($H'$ is not a group!!). So $S_3$ acts on $H'$ by conjugation. As you mentioned correctly $S_3$ contains 6 subgroups, where of course $S_3$, the trivial subgroup $\{1\}$ and the alternating group $A_3$ are three of them. The other three subgroups are easily found (and I guess you have written them down explicitly).

Now, you have to find the orbits of this action. So the "straightforward" check would be to take all six elements of $S_3$ and apply them to all of your six subgroups and see what happens. However you can shorten up a little bit, since you don't need any computation for the three subgroups you "know by name". Indeed, if you conjugate $S_3$ (as an element of $H'$) by any element of $S_3$ (the group which acts) you'll always obtain $S_3$. The same is true for $\{1\}$. Furthermore $A_3$ has index 2 in $S_3$, thus it is a normal subgroup. And, by definition, normal subgroups are exactly those subgroups which are fixed by conjugation. In other words conjugating $A_3$ (as an element of $H'$) with any element of $S_3$ (the group which acts) you 'll always get $A_3$ again. Thus, the orbit of any of this three subgroups (elements of $H'$) is just the subgroup itself.

However trying to do this for your remaining three elements of $H'$ (subgroups of $S_3$) this is not true.

For example you have the two subgroups $A=\{ 1, (13)\}$ and $B=\{ 1, (12)\}$. Is there really no element $g \in S_3$ such that $g A g^{-1} = B$? In total, taking this (easy) consideration to an end you should get that there are $4$ orbits (3 of them contain one element [see 2nd paragraph] and one contains three elements).

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  • $\begingroup$ so I got that the 4th orbit. I don't know what the meaning of Orbit-stabilizer theorem if H' is not a group. what do they want me to calculate? $\endgroup$ – user2993422 Sep 5 '15 at 18:02
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    $\begingroup$ If $G$ is a (finite) group acting on a (finite) set $X$ then for any $x \in X$ it holds that: $|G_x| \cdot |G(x)| = |G|$, where $G(x)$ is the orbit of $x$ (the $G$-translates of $x$) and $G_x$ is the stabilizer of $x$ (the set of all $g \in G$ which fix $x$). $\endgroup$ – M.U. Sep 5 '15 at 18:11
  • $\begingroup$ thanks a lot for the eplanation $\endgroup$ – user2993422 Sep 5 '15 at 20:35
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As you suspect, your answer is not correct. The first thing you need to list all the subgroups of $S_3$. Now for each subgroup $H \leq S_3$ and for each $g \in S_3$, you need to compute $gHg^{-1}$. These conjugate subgroups are the elements of the orbit of $H$.

For example, take $H = \langle (1 \ 2) \rangle \leq S_3$. Now we need to loop over all the $g \in S_3$ and compute $gHg^{-1}$. If we take $g = 1$ or $g = (1\ 2)$, we just get $gHg^{-1} = H$. Now consider $g = (1 \ 3)$. Since $$ (1\ 3) (1\ 2) (1\ 3)^{-1} = (1\ 3) (1\ 2) (1\ 3) = (2\ 3) $$ we see that $gHg^{-1} = \langle (2\ 3) \rangle$. Thus $H = \langle (1\ 2) \rangle$ and $\langle (2\ 3) \rangle$ are both in $O_H$, the orbit of $H$. What other subgroups can you get by conjugating $H$ by other $g \in S_3$? What happens for the other subgroups? Try $H = \langle (1\ 2\ 3) \rangle$ and see how this differs from $\langle (1\ 2) \rangle$.

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