Given a finite group and a prime $p$ which divides its order, Let $N(p)$ denote the number of $p-$sylow subgroups of $G$. If $G$ is a group of order $21$, what are the possible values of $N(3)$ and $N(7)$?

G has only $1$ $7$-Sylow subgroup (because this number must divide $3$ and be equal to $1$ mod $7$) and has either $1$ or $7$ ,$3$-Sylow subgroups (because this number must divide $7$ and be equal to $1$ mod $3$)

  • $\begingroup$ Both possibilities for the Sylow $\;3-$ are possible and actually happen: there's one unique such subgroup if the group is the only abelian one (the cyclic one), and 7 if it is not (the non-trivial semidirect product of the only Sylow $\;7-$subgroup and on of the seven Sylow $\;3-$subgroups). $\endgroup$
    – DonAntonio
    Jan 26 '14 at 6:03

There are two types of Group of order $21$.

$1$.$~~~\mathbb Z_{21}$

$2$.$~~~\mathbb Z_{7}\rtimes\mathbb Z_{3}$.

Let $G\cong \mathbb Z_{21}$. Then $G$ has exactly one subgroup of order $3$ and exactly one subgroup of order $7$ as, $G$ is a cyclic group. Thus $N(3)=N(7)=1$.

Assume that $G \cong \mathbb Z_{7}\rtimes\mathbb Z_{3}$. As you know $N(7)=1$. Now, we compute $N(3)$. By Sylow Theorem, we have $1+3k=1$ or $1+3k=7$. If $1+3k=1$, then $G$ is an abelian group and it is contradiction. Hence $N(3)=7$.


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