I am trying to solve the following exercise:

¿How many Sylow subgroups has a non abelian group $G$ of order $21$ and $39$ respectively.

I could do the following:

a) $|G|=21=3\cdot 7$. I'll call $n_p$ the number of $p-$Sylow subgroups. Then we have $$n_3 \equiv 1 (3), \space n_3|7 \implies n_3 \in \{1,7\}$$ $$n_7 \equiv 1 (7), \space n_7|3 \implies n_7=1$$

I have to determine if $n_3=1$ or $n_3=7$. I suppose that here is where I must use that $G$ is not abelian. I couldn't do much from here, all I could think of was: let $H$ be the $7$-Sylow subgroup and $K$ be a $3-$Sylow subgroup (we don't know if there are $7$ or just $1$ but at least there exists one). Then $H \cap K=\{1\}$. Since $H$ is normal, we have that $HK$ is a subgroup in $G$ and $|HK|=\dfrac{|H||K|}{|H \cap K|}=3.7=21$. From here it follows $G$ can be expressed as a semidirect intern product of$H$ and $K$, $G=H._sK$.

I did an analogous thing for the case $G$ is a group of order $39$ and got stuck at the same point. Could this help me to determine if $n_3=1$ or $n_3=7$? I would appreciate some help to finish the problem. Thanks in advance.


If $n_3=1$, the $3$-Sylow subgroup $H_3$ is normal. Since $H_7$, the $7$-Sylow subgroup, is also normal, it can be shown that these two subgroups commute, meaning that for every $a\in H_3$ and $b\in H_7$ $ab=ba.$ It follows that $G$ is a direct product, thus abelian.

To show that they commute, let $a\in H_3,b\in H_7$, and observe the commutator $aba^{-1}b^{-1}.$ Since both subgroups are normal, the commutator is in both of them, thus is equal to $1$.


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