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I am trying to show the following:

Prove that there are no simple groups of the given order:

42.

200.

231.

255.

I understand that they need to be broken down into their prime factors.

I was looking to try to understand: https://www.math3ma.com/blog/4-ways-to-show-a-group-is-not-simple and also what they do in No simple groups of order $90$ and No simple group of order 2016

What I don't understand, and from all the other problems, is how they determine $n_p\in\{...\}$. If I can understand this step I believe that I will be able to continue with the problem.

What I got for 42 is that the prime factors of $|G|$ are 2,3, and 7. Let $n_p$ denote the number of Sylow $p$-subgroups of $G$, this gives us that $n_2\in\{1,3,7,21\}$, $n_3\in\{1,2,14\}$, and $n_7\in\{1,6\}$. Suppose that none of these are 1, (if they are we are done because then there is another normal group, which by definition means that $G$ cannot be simple.) Consider $n_7=6$. This means that there are $7(6-1)=35$ elements of order 7.

I guess I would like to know if I am doing this right, or if there is some fact that I am missing.

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    $\begingroup$ You also have $n_p \equiv 1 \pmod{p}$ (see Theorem 3 here), which further limits possibilities. For instance in your order $42$ example, $n_7 \equiv 1 \pmod{7}$ so it can't be $6$. (And similarly, $n_3$ can't be $14$.) $\endgroup$ – André 3000 Apr 12 at 20:27
  • $\begingroup$ So does that change $n_7\in\{1,7,14,21\}$? $\endgroup$ – Nicole Havens Apr 12 at 20:29
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    $\begingroup$ $n_7$ must divide $6$ and be $\equiv 1 \pmod{7}$. So the only possibility is $n_7 = 1$. $\endgroup$ – André 3000 Apr 12 at 20:31
  • $\begingroup$ Why must it divide 6? $\endgroup$ – Nicole Havens Apr 12 at 20:32
  • $\begingroup$ This is the first bullet in the link I gave. A Sylow $7$ subgroup will have order 7, so it will have index $6$, and $n_p$ must divide this index. $\endgroup$ – André 3000 Apr 12 at 20:34

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