Given $H = \{ id , (12) , (34) , (12)(34)\}$ find $Stab(H)$ and the bijection from the set of left cosets of $Stab(H) $ in $ S_4 $ to $ \mathcal{O} (H) $ ; $S_4$ acts on $H$ by conjugation.
ie: $$\varphi: \frac{S_4}{Stab(H)} \to \mathcal{O} (H) $$
so $Stab(H) = \{ \sigma \in S_4 | \sigma H \sigma^{-1} = H \}$
my trouble is how do you find a $\sigma$ to act on $H$ when $H$ has elements of different cycle composition? since if $\sigma \pi \sigma^{-1} = \pi \ ; \ \sigma = (\sigma(\pi(1) )\sigma(\pi(2) )…\sigma(\pi(n) )$
the only solution I can think of is that $Stab(H) = H$
$\varphi:\frac{S_4}{H} \to \mathcal{O} (H) $
$\varphi(gh) = gh$
Best Answer
I presume you mean the conjugation action of $G=S_4$ on the set of its subgroups. In that case, the stabilizer of $H$ is $N_G(H)$, the normalizer of $H$.
You are correct that every element of $H$ stabilizes $H$. In fact, since $H$ has order $4$, it is abelian, so conjugation by an element $h\in H$ not only normalizes $H$, it centralizes $H$ (every element of $H$ is unchanged by conjugation by $h$).
But there are additional elements of $S_4$ which normalize $H$. I claim that we can find an element $g \in S_4$ such that conjugation by $g$ exchanges $(12)$ and $(34)$ while fixing $id$ and $(12)(34)$. Then $g$ normalizes $H$ but does not centralize $H$, so $g \not\in H$.
What element $g$ can we use to exchange $(12)$ and $(34)$? For example, $g=(13)(24)$ will accomplish this: check that $$g(12)g^{-1} = (34)$$ and $$g(34)g^{-1} = (12)$$ Of course $g(id)g^{-1} = id$, and finally, $$g(12)(34)g^{-1} = g(12)g^{-1}g(34)g^{-1} = (34)(12) = (12)(34)$$ So now we have at least five elements of $G$ which normalize $H$, namely the four elements in $H$, plus the element $g$ defined above. Since the normalizer is a subgroup of $G$, its order must divide $|S_4| = 24$. Since $5$ does not divide $24$, that means the normalizer contains additional elements that we have not found yet. See if you can find the rest.
Another way to see that the stabilizer (normalizer) of $H$ is more than just $H$ is to look at the orbit of $H$ under conjugation. See if you can recognize that these three subgroups are all conjugate to each other: $$\{id, (12), (34), (12)(34)\} = H$$ $$\{id, (13), (24), (13)(24)\}$$ $$\{id, (14), (23), (14)(23)\}$$ and since conjugation preserves cycle structure, these are the only conjugates of $H$. This means that the orbit of $H$ has size $3$. By the orbit-stabilizer theorem, the size of the orbit of $H$ is the same as the index of its stabilizer (normalizer). Therefore, $|G : N_G(H)| = 3$, which means that $|N_G(H)| = 8$. Above, we identified $5$ elements of $N_G(H)$, so this argument shows that there are $3$ more.