Let $G＝S_5$ be symmetric group. Let $H=\langle(1,2,3)\rangle$ be subgroup of $G$.
How many subgroups of $G$ which is conjugate to $H$?
My try: $H$ is sylow $3$ subgroup of $G$, so the number of conjugate subgroups are $1$ or $4$ or $10$ because it decides $5!÷３＝40$ and $\equiv 1\bmod 3$. And $H$ is not normal subgroup, so the number of conjugate subgroups of $H$ is $4$ or $10$. But I cannot proceed from here. Thank you in advance.
Any subgroup of $S_5$ which is conjugate to $H$ will be of the form $\langle(a,b,c)\rangle$, for some subset $\{a,b,c\}\subseteq \{1,2,3,4,5\}$ of size $3$. There are ${5\choose 3}=10$ such subsets. Note that $\langle(a,b,c)\rangle=\langle(c,b,a)\rangle$, so there is at most a unique subgroup conjugate to $H$ for each of these $10$ subsets.
Finally note that for any subset $\{a,b,c\}\subseteq \{1,2,3,4,5\}$, there is a permutation in $S_5$ taking the set $\{1,2,3\}$ to $\{a,b,c\}$. This element will conjugate $H$ to $\langle(a,b,c)\rangle$.
Thus all $10$ of the subsets correspond to subgroups conjugate to $H$.