What exactly is the only Sylow $3$-Subgroup of $D_3$?

Question

I can apply the needed theorem to get me to the fact that it only has one Sylow $3$-Subgroup but I donʻt know how to find exactly what it is. I have the multiplication table computed so help in regards using the table specifically would be beneficial.

Answer 1

There is exactly one group of order three up to isomorphism. Since $3$ is prime, $3\mid 6$, and $6$ is the order of $D_3$, Cauchy's Theorem gives us at least one element $r$ of $D_3$ of order three. It then follows that your subgroup is $\{e, r, r^2\}$.

Answer 2

Find a subgroup of order $3$. This is a Sylow subgroup. Like you said, there is only one. It is generated by an element of order $3$. Find such an element.

Answer 3

Actually, as an alternative to the other answers, note that $D_3\cong S_3$. Since the Sylow $3$ subgroup has order $3$, and $(123)$ has order $3$, it is $\langle (123)\rangle$.

This must be, under an isomorphism, the same as the subgroup generated by a rotation through $2\pi/3$. Namely $\langle r\rangle=\{e,r,r^2\}$.