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I recently came across a theorem that states that if G is a finite group and every p-group of G is normal then G is isomorphic to the direct product of its Sylow p subgroups. To prove this we use that in this case elements from different Sylow subgroups commute. Now in the abelian case (ie the $\mathbb{Z}$ module case) every subgroup is a direct summand of G by the classification theorem.

So I was wondering if we could do something similar in the non abelian case using normality. My question is:

If G is a finite group and P is a normal Sylow p subgroup then does there exist $H\leq G$ such that $G\cong P\times H$.

I believe this statement is not true in general (but it is true if all of the Sylow p subgroups are normal iff G is nilpotent), so what are some conditions for this to hold. Also what if we replaced the direct with semidirect product ?

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    $\begingroup$ There is always a subgroup $H$ such that $G = P \rtimes H$ by the Schur-Zassenhaus Theorem $\endgroup$
    – Derek Holt
    Apr 25 at 7:51
  • $\begingroup$ So to find conditions to get direct product we should look for conditions to make the semidirect product direct (also do you have any references for the proof of this theorem) $\endgroup$
    – Bigalos
    Apr 25 at 8:07
  • $\begingroup$ It's a direct product if and only if the homomorphism defining the semidirect product is trivial I don't think there is much more to say. $\endgroup$
    – Derek Holt
    Apr 25 at 8:32
  • $\begingroup$ "In the abelian case, every subgroup is a direct summand of $G$ by the classification theorem." This is false. The subgroup of order $2$ is not a direct summand of the cyclic group of order $4$. $\endgroup$
    – Arturo Magidin
    Apr 25 at 18:24
  • $\begingroup$ A trivial counterexample to your question is $S_3$, where the Sylow $3$-subgroup is normal, but it cannot be written as $S_3=A_3\times C_2$. $\endgroup$
    – Arturo Magidin
    Apr 25 at 18:25

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