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?

  • 1
    $\begingroup$ I would say that $n_{3}|8$. So $n_{3}$=1 or $n_{3}$=4. $\endgroup$ – nilcorc Dec 3 '14 at 21:20
  • $\begingroup$ More precisely, the number of $3$-Sylow subgroups divides $\frac{72}{9} = 8$. $\endgroup$ – Daniel Fischer Dec 3 '14 at 21:20
  • $\begingroup$ $n_3\mid 8$ is of course not stronger than $n_3\mid 24$ as $n_3\equiv 1\pmod 3$. $\endgroup$ – Hagen von Eitzen Dec 3 '14 at 21:23

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 $n_3=4$?

  • $\begingroup$ As the question posted by the OP says "... can ....have", I think that what he did is enough. $\endgroup$ – Timbuc Dec 3 '14 at 21:40

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