Show that $|G|=1000$ is not simple

Question

I think I've done this, but I just want to check it's proved enough.

So $|G|=1000=8\cdot125$.

Then $n_5\in\{\text{factors of 8}\}$ so $n_5\in\{1,2,4,8\}$ congruent to $1\bmod5$.

$2$, $4$, $8$ are not congruent to $1\bmod5$.

So $n_5=1$ and if $n_p=1$ then the Sylow $p$-subgroup is unique and normal so $G$ cannot be simple.

Yes, and for the general idea see http://math.stackexchange.com/questions/13111/no-simple-group-of-order-300?rq=1. — Dietrich Burde, Dec 10 '13 at 19:42
Everything is fine, you could add the proof verification tag :-) — Fakemistake, Sep 18 '19 at 17:42

score -1 · Answer 1 · answered Aug 07 '23 at 07:19

-1

In general, we know that every group whose order is divisible by exactly two primes is not only non-simple, but also solvable.

answered Aug 07 '23 at 07:19

Ash

1

It's probably worth mentioning the name of the result, https://en.wikipedia.org/wiki/Burnside%27s_theorem – Travis Willse Aug 07 '23 at 07:50
Thanks for mentioning that. – Ash Aug 07 '23 at 07:54

1 Answers1