Today I've been looking the third Sylow theorem. My professor did an example in class, so I tried to solve the example by myself and then compare what I did with the answer of my professor. The exercise is this:
Let $G$ be a group, and $\lvert G\rvert=6$. Find the number of Sylow $2$-subgroups and Sylow $3$-subgroups.
This is my solution:
We see that $\lvert G\rvert=6=2\cdot 3.$ Let: \begin{equation} \begin{cases} n_2=\text{number of Sylow $2$-subgroups}\\ n_3=\text{number of Sylow $3$-subgroups} \end{cases} \end{equation} By the third Sylow theorem, we know that: \begin{equation} \begin{cases} n_2 \mid 3\ \ \text{and}\ \ n_2\equiv 1\ \ \text{(mod}\ 2)\\ n_3 \mid 2\ \ \text{and}\ \ n_3\equiv 1\ \ \text{(mod}\ 3)\\ \end{cases}, \end{equation} so \begin{equation} \begin{cases} n_2=1\ \text{or}\ 3\\ n_3=1\\ \end{cases} \end{equation}
But my professor wrote this:
\begin{equation} \begin{cases} n_2=1\\ n_3=1\ \text{or}\ 3\\ \end{cases} \end{equation} And we see that \begin{equation} n_3= \begin{cases} 1\to \text{This Sylow $3$-subgroup is normal}\\ 3\to \text{Those three subgroups are conjugate between them} \end{cases} \end{equation}
I don't see why he says that $n_3=1$ or $n_3=3$. By the third Sylow theorem, the only possibility that I see for $n_3$ is $1$.
You're absolutely right that $n_{3} = 1$.
It's $n_{2}$ that can be either $1$ (in a cyclic group) or $3$ (in $S_{3}$).