Proof Verification: A group of order $595$ has a normal Sylow 17-subgroup.
$|G|=595=5.7.17$
The divisors of $595$ are $1,5,7,17,35,85,119,595$.
$17|n_{17}-1\implies n_{17}=1,35\\7|n_7-1\implies n_7=1,85,119\\5|n_5-1\implies n_5=1$
If possible let $n_{17}=35.$ Then there is at least $35.17-34=561$ element of order $17.$ Then $n_7=1$ for otherwise the number of elements of order $17,7$ will exceed $595!$
Let $H_5,H_7$ be the normal Sylow 5 and Sylow $7$ subgroups of $G.$ Then $H_5H_7\le G$ and since $H_5\cap H_7=(e),G=H_5\times H_7\simeq H_5\oplus H_7.$ Thus none of the $35$ elements in $H_5H_7$ is of order $17.$ Hence $G$ has at least $561+35=596$ elements!
Please tell me whether the proof is right?
Best Answer
I'd try as follows:
Either $\;H_7\lhd G\;\;or\;\;H_{17}\lhd G\;$, otherwise we'd have at least $\;35\cdot 16+ 85\cdot 6=1,070 \;$ different elements in $\;G\;$ and that's impossible.
Assume thus that
$$\;H_7\lhd G\;\implies H_7H_{17}\;\;\text{is a subgroup of index $\;5\;$ in}\;\;G$$
and since $\;5\;$ is the minimal prime that divides $\;|G|\;$ we get that $\;H_7H_{17}\lhd G\;$
But then we're done since $\;H_7H_{17}\;$ is a cyclic group with an obviously normal subgroup of order $\;17\;$, and normal subgroup of normal cyclic subgroup is normal itself, i.e.
$$A\lhd B\lhd G\;,\;and\;\;B\;\;\text{cyclic}\;\implies A\lhd G$$