= Solution
Since $351=3^3\cdot13$, the number of <Sylow subgroups> for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper <normal subgroup>.
In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every <Sylow subgroup> for $p=3$ has order $27$, and all its elements lie among these remaining elements, by <Lagrange's theorem>. It must therefore be precisely that set. The <Sylow subgroup> for $p=3$ is unique and hence <normal>. \b[In either case $G$ is not a <simple group>.]
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