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By the "$2n$-test", proving that a group of order $210$ cannot be simple. Is there another way to prove this? Would you use Sylow Theory?

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user137957
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  • OK, it became personal after my last blown answer :) It seems to me this idea should work, but I welcome comments and corrections: By element-counting and group action stuff, the critical case is that in which there are 15 7-Sylows, 21 5-Sylows, 10 3-Sylows and $\le 15$ 2-Sylows. Look at a 7-Sylow. Its normalizer has order 14, and it can't be $D_7$, because if it were we would end up with too many $2$-Sylows. So we have 15 copies of $Z_{14}$ in $G$, and they all intersect trivially. Now we get a contradiction by counting elements. – John Brevik Apr 13 '15 at 17:18
  • I would say that since the normalizer of a 7-Sylow has order 14, there cannot be a subgroup isomorphic to $Z_{21}$ since that would mean the normalizer of the 7-Sylow would have order equal or greater than 21. This means the normalizer of a 3-Sylow, which has order 21, has to be the non-Abelian group of order 21, which I sometimes call C(3,7). The 3-Sylow is thus normal in C(3,7) contradicting the non-normality of subgroups of order 3 in C(3,7). So no simple group of order 210. – jimvb13 Jan 23 '19 at 00:22
  • Another way: For every prime $p$, if Sylow $p$-subgroup of $G$ is cyclic, then $G$ is solvable. This Theorem yells that every group of order $p_{1}p_{2}…p{n}$, where $p_{i}\ne p_{j}$ is a prime, is solvable, for instance a group of order $2\cdot 3 \cdot 5 \cdot 7=210$. – Ash Aug 19 '23 at 21:52

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To quote a comment to a question about groups of order 105:

A finite group whose order $n$ is not divisible by the square of any prime (or more generally has no chief factor whose order is divisible by the square of any prime) has subgroups of order $d$ for every divisor $d$ of $n$, but those subgroups don't have to be cyclic (though in the square-free case they are all built from at most two cyclic subgroups, so you might want to look more carefully at the possible groups of order 21 to see the full story isn't much more complicated). – Jack Schmidt

In particular, this means that any such group has a Hall $\pi$-subgroup for any set of primes $\pi$, and is therefore solvable and so not simple. Your group is such a group.

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