How many Sylow-$ 3$ subgroup does $G$ have? Let $G$ be a noncyclic group of order $21$.  How many sylow-$3$ subgroup does G have?
The possible orders of Sylow $3$ subgroups is $1, 7$. But how to check the exact number?

By the way, you can realize this group concretely in the following form. Let $F = \mathbb{Z}/7\mathbb{Z} = \{0, 1, \dots, 6 \}$ be the field with $7$ elements, and
$$
G = \left\{ \begin{bmatrix} 1 & a\\0 & b\end{bmatrix} :
a \in F, b \in \{1, 2, 4 \} \right\}.
$$
Note that $\{1, 2, 4 \}$ is a subgroup of $F^{\star}$ of order $3$.
You can verify that
$$
G = \left\{ \begin{bmatrix} 1 & a\\0 & 1\end{bmatrix} :
a \in F \right\}
$$ 
is the $7$-Sylow subgroup, and that all elements outside it have order $3$. This is because
$$
\begin{bmatrix} 1 & a\\0 & b\end{bmatrix}^{3}
=
\begin{bmatrix} 1 & a(1+b+b^{2})\\0 & b^{3}\end{bmatrix},
$$
and for $b \in \{ 2, 4 \}$ we have $b^{2} + b + 1 = 0$ in $F$ (and clearly $b^{3} = 1$).
