How many Sylow $3$-subgroups can a group of order $72$ have?
Let $G$ be a group of order $72=2^3 \cdot 3^2$. The number of Sylow $3$-subgroups $n_3$ divides 24 and has the form $n_3=3k+1$ by the Sylow Theorems. Therefore $n_3=1$ or $n_3=4$.
Am I done?
Best Answer
To complete the task, you should also show that $1$ and $4$ are indeed possible. That is, exhibit examples of groups with these counts: $n_3=1$ is witnessed by $\mathbb Z/72\mathbb Z$. Can you name a group with $n_3=4$?