# A group of order $595$ has a normal Sylow 17-subgroup.

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?

• Well, certainly $\;H_5H_7\cong H_x\times H_7\;$ as they both are normal and thus generate their own direct product, but of course $\;G\neq H_5H_7\;$ ... Dec 11, 2013 at 5:28
• Does this answer your question? Prove that every group of order $595$ is cyclic.
– user1007416
Aug 30, 2022 at 11:00

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.

$$A\lhd B\lhd G\;,\;and\;\;B\;\;\text{cyclic}\;\implies A\lhd G$$

you have count 1 twice. So your proof is not right.

• Can you please explain? DonAntonio calculated $35(17-1)+85(7-1)$, so this is correct. Dec 11, 2013 at 16:38
• Of course, DonAntonoio is right. My answer is for the proof of user113578. Dec 11, 2013 at 23:50