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How can I show that a group with 380 elements is not simple?

Vinícius
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    In fact, groups of order $p(p+1)$, for $p$ a prime, always have a normal subgroup, of order $p$ or order $p+1$. –  Jul 09 '12 at 21:03

2 Answers2

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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?

Arturo Magidin
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Can we also take the approach that a $5$-Sylow subgroup would have order $5$ and if there are greater than $1$ of them, there will be $76$, hence $(5-1)\cdot76=304$ elements of order $5$. Then, the number of $19$-Sylow subgroups must be $1$, since otherwise there are $20$ and that’s too many elements.

Mc5231
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