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 $G$.
$\\\\$
I have no idea how to prove this in either direction, so any help is appreciated.
2
$\begingroup$
$\endgroup$
-
$\begingroup$ See this question. $\endgroup$ – Dietrich Burde Mar 5 '18 at 16:17
1
$\begingroup$
$\endgroup$
$(\Rightarrow)$ 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 $P\triangleleft G$.
0
$\begingroup$
$\endgroup$
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.
-
$\begingroup$ How do you use $P$ being normal when you prove $\Rightarrow$ ? $\endgroup$ – Levent Mar 5 '18 at 16:27
-
1$\begingroup$ Can you clarify? E.g., $S_3$ is generated by elements of order $2$. $\endgroup$ – John Brevik Mar 5 '18 at 16:29
-
$\begingroup$ Yes, I am sorry in my proof I assumed $G$ abelian. Let me change my answer. $\endgroup$ – user152715 Mar 5 '18 at 16:34
-
-
