I want to find out if a group $G$ of order $1000$ contains a proper normal subgroup $H$.
- Can one solve this without the use of Sylow theorems?
- Let's assume we will use the theorems of Sylow. We write $1000=2^3 5^3.$ For $q=5$,we find that the number of $q$-Sylow subgroups, $n_q,$ has only the solution $n_q=1.$ That means that $G$ contains a normal sungroup $H$ of order $125.$ I have nevertheless a problem with $p=2.$ In that case I get $n_p\in \{ 1,5,25,125\},$ meaning that there are many solutions for $n_p$. What does this mean i.e. how many $p$-Sylow subgroups are there in $G$? Thanks for any comment.
Best Answer
As you find yourself, $G$ surjects onto a group of order 8 with kernel order 125. Both are prime-power order, therefore nilpotent and solvable, hence so is $G$.
As to the latter, let $D_{10}$ be the dihedral group of order 10　and $C_n$ be the cyclic group of order $n$. The former has five 2-Sylow subgroups, while the latter has one. It follows that $C_{1000}, C_{100}\times D_{10}, C_{10}\times D_{10}\times D_{10},D_{10}\times D_{10}\times D_{10}$ have 1,5,25,125 2-Sylows respectively.