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For finite group $G$ and Sylow p-subgroup $P$, prove that

$P \unlhd G \iff$ all subgroups generated by elements of order a power of $p$ are p-subgroups of $G$.

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I have no idea how to prove this in either direction, so any help is appreciated.

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2 Answers 2

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$(\Rightarrow)$ part is easy since every $p$-power element of $G$ is in $P$, any subgroup generated by $p$-power elements is contained in $P$.

$(\Leftarrow)$ is again easy. Say there are two Sylow subgroups $P,Q$. Let $S=P\cup Q$. Then $S$ generates a $p$-group containing both $P$ and $Q$. Since $P,Q$ are Sylow, that implies $S=P=Q$ so $P\triangleleft G$.

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If this is you definition of $p$ subgroup then

$\Rightarrow$

Subgroups generated by elements having order $p^k$ form are subgroups of $P$.

claim: $P$ is a $p$ group iff $|P|=p^n$

Pf: $\Leftarrow$ is trivial.

Conversely, suppose $\exists q||P|$ then by cauchy's theorem there will be an element of order $q$ in $P$. Contradiction. Hence the claim.

Using that claim we get that those subgroups are $p$ groups.

$\Leftarrow$,

$P$ is normal as it is fixed by each conjugation operation.

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  • $\begingroup$ How do you use $P$ being normal when you prove $\Rightarrow$ ? $\endgroup$
    – Levent
    Mar 5, 2018 at 16:27
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    $\begingroup$ Can you clarify? E.g., $S_3$ is generated by elements of order $2$. $\endgroup$ Mar 5, 2018 at 16:29
  • $\begingroup$ Yes, I am sorry in my proof I assumed $G$ abelian. Let me change my answer. $\endgroup$
    – Ri-Li
    Mar 5, 2018 at 16:34
  • $\begingroup$ Thanks, both @JohnBrevik . $\endgroup$
    – Ri-Li
    Mar 5, 2018 at 16:35
  • $\begingroup$ Thanks, @Levent $\endgroup$
    – Ri-Li
    Mar 5, 2018 at 16:36

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