If $G$ is a group of order $231$, prove that the $11$-Sylow subgroup is in the center of $G$.
My efforts: Since $o(G)=231=3\times 7\times 11$ then using third Sylow's theorem we get the following result: $N_{11}=N_{7}=1$ and $N_3=1 \ \text{or}\ 7$, where $N_p$ - the number of Sylow $p$-subgroups.
Let $H_7, \ H_{11}$ be Sylow $7$ and $11$-subgroups. Then it is easy to show that $H_7\cap H_{11}=\{e\}$ and $H_7, H_{11}$ are normal in $G$. Thus, $H_7H_{11}$ is an abelian subgroup of order $77$ in $G$.
The following inclusion is true: $Z_G\subset Z_{H_7H_{11}}=H_7H_{11}$ so $Z_G|77$ $\Rightarrow$ $|Z_G|\in \{1,7,11,77\}$.
Suppose that $N_3=1$ then $H_3$ is normal in $G$ and $H_3H_7$ is an abelian subgroup of order $21$. Applying the same method in that case we got the following options: $|Z_G|\in \{1,3,7,21\}$. Combining this one with above we get that $|Z_G|=1, 7$. However in both cases, Sylow $11$-subgroup cannot lie in the $Z_G$.
Is this approach correct?
I have met this problem in this forum but its solution was different. I would like to complete my own solution but I am not able to do it with the case $N_3=7$.
Would be grateful for elementary solution.
I don't understand that part. What if, say, $G$ was $\mathbb{Z}_{231}$? Then $Z_G$ would be $G$ itself and certainly not included in $Z_{H_{11}H_7}$... right?
What is true using your proof below is that $$(H\cap Z_G) \subset Z_H$$ whenever $H$ is a subgroup of $G$.
Another approach:
$G$ acts by conjugation on $H_{11}$ because there is only one $11$-Sylow as you have proved. What you want is to show that this action is trivial.
Since $H_{11}$ is cyclic it is abelian, and therefore $H_{11}$ lies in the kernel of the action morphism $$G\rightarrow \operatorname{Aut}(H_{11})$$
Hence this morphism factors through $G/H_{11}$. But now $G/H_{11}$ is of order $21$ and $\operatorname{Aut}(H_{11})$ is of order $10$, therefore there are no non-trivial morphisms $$G/H_{11}\rightarrow \operatorname{Aut}(H_{11})$$ QED.