D+F 6.2.3

$|G|=380 \Longrightarrow $ $G$ is not simple.

$\textit{Proof}$ : First note that $380=2^2 \cdot 5 \cdot 19$. By Sylow's theorem (iii), we get that for $p=19$, $n_{19}=1$(mod 19) and also $n_{19} \mid 2^2 \cdot 5 = 20$. So, the possible values of $n_{19}$ are $1,20$.
For $p=5$, we get that $n_5=1$(mod 5) and also $n_5 \mid 2^2 \cdot 19 = 76$. Given these two constraints, the possible values for $n_5$ are $n_5=1,76$.
For $G$ to be simple, we cannot have any non-trivial normal subgroups. This means that by corollary 20 p.142, we must have $n_p=1$ for all $p$ in the factorization of $|G|$.

Suppose that it is the case that $n_{19}=20$ and $n_5=76$. The Sylow-19 subgroups have order 19, (the 19 factor in 380 is square-free) and the only group of order 19 is $\mathbb{Z}_{19}$, which has 18 elements of order 19. With $n_{19}=20$, this would mean that our entire group $G$ has $20 \cdot 18=360$ elements of order 19. The Sylow 5-subgroups have order 5, and the only group of order 5 is $\mathbb{Z}_5$, which has 4 elements of order 5. With $n_5=76$, this means our entire group $G$ would have $76 \cdot 4 = 304$ elements of order 5. This is an immediate contradiction on the size of $G$. Thus, it must be the case that either $n_5=1$ or $n_{19}=1$. In each case, we would have a non-trivial normal subgroup making $G$ not simple. $\blacksquare$