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Show that any group of order $20$ is not simple.

Denote the group $G$. It seems intuitive to state first that $20=2^2 \times 5$. Sylow's theorem then states that since a prime number, $5$, divides the order of $G$, we must have a subgroup of $G$ of order $5$. Where do I go from here?

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    $\begingroup$ The standard thing to do from here is to use Sylow 3 to figure out whether one of the Sylow subgroups has to be normal. $\endgroup$ Apr 26, 2016 at 19:01

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By Sylow, the number $n_5$ of Sylow $5$-subgroups of $G$ divides $4$ and satisfies $$n_5 \equiv 1 \mod 5.$$ Therefore we must have $n_5 = 1$, but this is equivalent to saying that the Sylow $5$-subgroup is normal.

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