Group of order 63 I googled my question, nothing appeared. My book says that group of order 63 is Abelian. The way I see it is perfectly possible that it has 7 Sylow 3 subgroups and one Sylow 7 subgroup. Please help!
 A: If that is the case, what are the (maximal) numbers of elements of orders


*

*is either $3$ or $9$?

*is equal to $7$?

A: I'll try to write an answer describing groups of order $63=3^2\cdot7$.
(If there is any mistake please let me know)
It is $n_7|9,\ n_7\equiv 1\pmod7\Rightarrow n_7=1$ so there is a unique Sylow $7-$subgroup $P_7\lhd G$.
It is also $n_3|7,\ n_3\equiv 1\pmod3\Rightarrow n_3=1,7$. Let $P_3$ be a Sylow $3-$subgroup. It is $P_3=9$ so $P_3\cong \mathbb{Z}_9$ or $\mathbb{Z}_3\times\mathbb{Z}_3$

*

*If $n_3=1$ then $P_3\lhd G$ and $|P_3P_7|=|G|$ with $P_3P_7\leq G$ hence $G=P_3\times P_7\cong \mathbb{Z}_9\times \mathbb{Z}_7$ or $\mathbb{Z}_3\times \mathbb{Z}_3\times \mathbb{Z}_7$.

*If $n_3=7$ it is $G=P_7\rtimes_{\phi}P_3$ where $\phi:P_3\to Aut(P_7)$. Since $P_3\not\lhd G$ it follows that $\phi$ is non-trivial. Hence (since $Aut(\mathbb{Z}_7)\cong \mathbb{Z}_6$) $\phi$ maps $P_3$ to the unique subgroup of $\mathbb{Z}_6$ of order $3$. So $G\cong \mathbb{Z}_7\rtimes_{\phi}\mathbb{Z}_9$ or $G\cong \mathbb{Z}_7\rtimes_{\phi}[\mathbb{Z}_3\times \mathbb{Z}_3]$.

In the first case $\phi:P_3=\langle x\rangle \to Aut(P_7)=\mathbb{Z}_6=\langle \tau\rangle,\phi(x)=\tau,\ \tau(a)=a^2$ and in the second case $\phi:\langle x\rangle\times\langle y\rangle\to Aut(P_7)=\mathbb{Z}_6=\langle \tau\rangle,\ \phi(x)=\tau,\phi(y)=1,\tau(a)=a^2$
