With $G=S_{5}$ acting on $X$ by conjugation, prove $|X|=10$.

Question

Multiple problems, that don't all fit in the title.

Let $G=S_{5}$ and let $X=\{H \leq G: |H| =3 \}$. It is given that $G$ acts on $X$ by conjugation, so by mapping $(g,H)$ in $G \times X$ to $gHg^{-1}= \{ ghg^{-1} \text{ with } h \text{ in } H \}$ in $X$.

Problem 1. Prove $|X|=10$ and that all elements of $X$ are in the same orbit for this action.

Problem 2. Let $\sigma=(123)$ and $H= \langle \sigma \rangle$. Explain why $F=N_{G}(H)$ is a subgroup with $12$ elements and give $F$ in the form $\langle f_{1}, f_{2} \rangle$.

First of all, I find it hard to even understand what exactly is asked. Since $|H|$ has only three elements and is a subgroup of $G$, I would argue that one of the three elements in the identity, leaving only two options per subgroup $H$. Also I know there are exactly $20$ elements of order $3$ in $G$, namely all the permutations of the $3$ cycle (which has $\frac{5!}{3!}=20$ options).

This is all I have got, any hints are appreciated.

Answer 1

Indeed, there are $20$ elements of order $3$. Now, each subgroup of order $3$ contains exactly two elements of order $3$. More than that, any two distinct subgroups of order $3$ intersect trivially, this follows from Lagrange's theorem. So that way we get that $S_5$ has exactly $10$ subgroups of order $3$. Now why all such subgroups are in the same orbit? That is because here subgroups of order $3$ are actually $3$-Sylow subgroups of $S_5$, hence by Sylow theorems they are all conjugate to each other.

Answer 2

Hint for problem $2$: For the conjugation action of $G=S_5$ on $X$, use the orbit-stabilizer theorem. Note that $N_G(H)$ is the stabilizer of $H$.

Now, $N_G(H)=\langle (123), (234)\rangle $, as $(234)=(14)(123)(14)$ is a conjugate of $(123)$. (The subgroup $(123)$ and $(234)$ generate has order $12$.)