How can I show that a group with 380 elements is not simple?
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Note that $380=2^2\times 5 \times 19$. A Sylow subgroup associated to $19$ is necessarily cyclic of order $19$; if it is normal, we are done. And if it is not normal, then there must be twenty such subgroups; any two intersect trivially, since they are groups of prime order, so the twenty subgroups account for $20\times 18 + 1 = 361$ elements.
Now, consider the Sylow $5$-subgroups; how many can there be if there are twenty Sylow $19$-subgroups?