In particular, we will use Sylow’s theorem (3) and (4), and its corollary in the proof below.

For (b), recall that a group $G$ is solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such that the factor groups $G_i/G_{i-1}$ are all abelian groups for $i=1,2,\dots, n$.

Proof.

(a) The group $G$ has a normal Sylow $p$-subgroup

By Sylow’s theorem, the number $n_p$ of Sylow $p$-subgroups of $G$ satisfies $n_p\equiv 1 \pmod{p}$ and $n_p$ divides $q$.
The only such number is $n_p=1$.

Thus $G$ has the unique Sylow $p$-subgroup $P$ of order $p$.
Since $P$ is the unique Sylow $p$-subgroup, it is a normal subgroup of $G$.

(b) The group $G$ is solvable

Let $P$ be the normal Sylow subgroup of $G$ obtained in (a).
Then we have the following subnormal series
\[\{e\} \triangleleft P \triangleleft G,\]
where $e$ is the identity element of $G$.

The factor groups are $G/P$ and $P/\{e\}\cong P$.
The order of the group $P$ is the prime $p$, and hence $P$ is an abelian group.
The order $|G/P|=|G|/|P|=pq/q=q$ is also a prime, and thus $G/P$ is an abelian group.
Thus the factor groups are abelian. Thus $G$ is a solvable group.

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Hint.
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Use Sylow's theorem. To review Sylow's theorem, check […]

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Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$.
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