Note the prime factorization $217=7\cdot 31$.
We first determine the number $n_p$ of Sylow $p$-group for $p=7, 31$.
Recall from Sylow’s theorem that
\begin{align*}
&n_p \equiv 1 \pmod{p}\\[6pt]
&n_p \text{ divides } n/p.
\end{align*}

Thus, $n_7$ could be $1, 8, 15, 22, 29,\dots$ and $n_7$ needs to divide $217/7=31$.
Hence the only possible value for $n_7$ is $n_7=1$.
So there is a unique Sylow $7$-subgroup $P_7$ of $G$.

By Sylow’s theorem, the unique Sylow $7$-subgroup must be a normal subgroup of $G$.

Similarly, $n_{31}=1, 32, \dots$ and $n_{31}$ must divide $217/31=7$, and hence we must have $n_{31}=1$.
Thus $G$ has a unique normal Sylow $31$-subgroup $P_{31}$.

Note that these Sylow subgroup have prime order, and hence they are isomorphic to cyclic groups:
\[P_7\cong \Zmod{7} \text{ and } P_{31}\cong \Zmod{31}.\]

It is also straightforward to see that $P_7 \cap P_{31}=\{e\}$, where $e$ is the identity element in $G$.

In summary, we have

$P_7, P_{31}$ are normal subgroups of $G$.

$P_7 \cap P_{31}=\{e\}$.

$|P_7P_{31}|=|G|$.

These yields that $G$ is a direct product of $P_7$ and $P_{31}$, and we obtain
\[G=P_7\times P_{31}\cong \Zmod{7} \times \Zmod{31}\cong \Zmod{217}.\]
Hence $G$ is a cyclic group.

(b) Determine the number of generators of the group $G$.

Recall that the number of generators of a cyclic group of order $n$ is equal to the number of integers between $1$ and $n$ that are relatively prime to $n$.
Namely, the number of generators is equal to $\phi(n)$, where $\phi$ is the Euler totient function.

By part (a), we know that $G$ is a cyclic group of order $217$.
Thus, the number of generators of $G$ is
\begin{align*}
\phi(217)=\phi(7)\phi(31)=6\cdot 30=180,
\end{align*}
where the first equality follows since $\phi$ is multiplicative.

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