D+F 4.5.15

Prove that a group of order 351 has a normal Sylow p-subgroup for some prime $p$ dividing its order.

$|G|=351 \Longrightarrow \exists H \lhd G$ and $\exists p$ prime s.t. $H\in{Syl_p(G)}$

$\textit{Proof}$: Note that $351=3^3 \cdot 13$. Writing $|G|=351=p^am$, then part (iii) of Sylow's Theorem implies:
$n_p \equiv 1$(mod p) and $n_p \mid m$
First, look at $n_{13}$. Picking larger primes first is sometimes advantageous because there tends to be less possibilities when generating a list.
$351=13^1 \cdot 27$. So, $n_{13} \equiv 1$(mod 13) and $n_{13} \mid 27$
So, $n_{13}=1,27$.
If $G$ is simple, then $n_{13} \neq 1$ as this would make the Sylow 13-subgroup normal by Cor.20 p. 142. Thus, if $n_{13}=27$ this means that there are $27 \cdot (13-1) = 324$ elements of order 13. Since $|G|=351$, this leaves us with $351-324=27$ elements remaining. But the Sylow 3-subgroups have order $3^3=27$. So in this case this lack of freedom forces $n_3=1$. Thus, we have a non-trivial normal subgroup, making $G$ not simple. In any case, it must be that $n_p=1$ for $p=3$ or $p=13$.
$\therefore \quad G$ is not simple. $\blacksquare$

Note: this problem also solves one part of exercise 6.2.4