If every Sylow subgroup of a finite group $G$ is normal for every prime $p$, then $G$ is the direct product of its Sylow groups.

Question

I want to prove the following proposition.

If every Sylow subgroup of a finite group $G$ is normal for every prime $p$, then $G$ is the direct product of its Sylow groups.

IDEA.

I have the idea to use the following theorem.

THEOREM

If every Sylow subgroup of $G$ is a normal subgroup, then $G$ is isomorphic to the product of its Sylow subgroups.

One part of Sylow's theorem is that $G$ acts transitively by conjugation on its Sylow $p$-subgroups. This is very useful. — David A. Craven, Aug 08 '20 at 09:26
Hey, this question has already been answered here: https://math.stackexchange.com/questions/250941/g-is-the-product-of-its-sylow-subgroups — Raquel Magalhães, Sep 07 '20 at 21:16

score 0 · Answer 1 · answered Aug 08 '20 at 00:22

Let $L,M$ two different sylow groups. $x\in L, y\in M$, $xyx^{-1}y^{-1}\in L\cap N$ implies that $xyx^{-1}y^{-1}=1$ since the order of $L$ and $M$ are relatively prime. This implies that he subgroup generated by the Sylow subgroups is isomorphic to the product of the Sylow subgroups and its cardinal is the cardinal of $G$.

score 0 · Answer 2 · answered Aug 08 '20 at 09:19

0

Try proving that any group has only one Sylow - $p$ - subgroup $H$ if and only if $H$ is normal in $G$.

answered Aug 08 '20 at 09:19

abcd123

328

If every Sylow subgroup of a finite group $G$ is normal for every prime $p$, then $G$ is the direct product of its Sylow groups.

2 Answers2