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Let $G = S_4$, $H = S_3$, $X = G/H$ be the set of right cosets of $H$, $x = (14)H$ and $G $ acts on $X$ by conjugation. Compute $\mathscr{O} (x)$ and $G_x$ (the stabilizer of $x$).

I've got a problem in this question. Clearly, $\mathscr{O}(x) = \{(14)H, (24)H, (34)H, H \}$and $G_x = 1$, however it leads to a contradiction, since $|G_x| = |G|/|\mathscr{O}(x) | = 24 / 4 = 6$. I really do not know what's wrong in my answer.

Thanks in advance.

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up vote 4 down vote accepted

First I assume $H$ is permutations of $\{1, 2, 3\}$ and fixes $4$.

The subgroup $H$ is not normal in $G$, so the conjugation action of $G$ on itself does not descend to a conjugation action of $G$ on $G/H$. For example if we conjugate $H$ by $(1 \ 4)$ we get the copy of $S_3 \subseteq S_4$ given by permutations of $\{2, 3, 4\}$. But this subgroup is not a right coset of $H$.

If you want an action of $G$ on $G/H$ use left multiplication. Or, if you want your action to be conjugation you could act on the set of subgroups, or even the set of subsets of $G$.

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The action is given by $h .(gH) = h^{-1}ghH$. – user40276 Apr 30 '13 at 18:52
If $g \in H$ then $gH = 1H$. For the action to be well defined you would then need $hgh^{-1}H = h1h^{-1}H = 1H$, i.e., you would need $hgh^{-1} \in H$. But this is not true for all $g \in H$ because $H$ is not normal. – Jim Apr 30 '13 at 18:55
Sorry for the ambiguity, but I'm considering just representatives of each coset, then such $gH$ (with $g \in H$) would not be in $X$, therefore I think it's well defined. – user40276 May 1 '13 at 21:10
@user40276: If you want the action to be on elements then you are acting by conjugation on $G$, not on $G/H$. The conjugation action on $G$ is perfectly well defined, the conjugation action on $G/H$ is only well defined when $H$ is normal. – Jim May 1 '13 at 23:16
Sorry for my insistence, but this is not exactly about what I want. This question was written in a list of exercises that I received as a homework. So I'm trying to understand what was intention of the person that created this question. That's why I suggested the action on a set of representatives of each coset and, in this case, $g \notin X$. – user40276 May 2 '13 at 19:52

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