Suppose $G$ is a locally finite group such that $G=\bigcup\_{i=1}^\infty S\_i$, where $S\_i$ is a finite group and $S\_i \triangleleft S\_{i+1}$ for all $i \in \mathbb{N}$. Let $P$ be a Sylow (maximal with respect to inclusion) $p$-subgroup of the group $G$. For each $i \in \mathbb{N}$ put $P\_i=S\_i \cap P$. **Then $P\_i$ is a Sylow $p$-subgroup of $S\_i$ for all $i \in \mathbb{N}$. Or not?**



Sylow p-subgroups of locally finite groups
Yes. Fix $i$. For $n\ge i$, let $L\_n$ be a $p$-Sylow of $S\_n$ containing $P\_n$. Note that $S\_i$ is subnormal in $S\_n$. By the lemma below, $L\_n$, the intersection $M\_n=L\_n\cap S\_i$ is a $p$-Sylow of $S\_i$. Now there exists an infinite set $N$ of $n$ such that the Sylow $M=M\_n$ does not depend on $n\in N$. Since $\langle P\_n,M\rangle$ is a $p$-group for all $n$ and $P=\bigcup P\_n$, we deduce that $\langle P,M\rangle=\bigcup\langle P\_n,M\rangle$ (increasing union) is a $p$-group. By maximality, we deduce that $M\subset P$, QED.


[Lemma: if $G$ is a finite group, $H$ a subnormal subgroup and $P$ a $p$-Sylow of $G$ then $P\cap H$ is a $p$-Sylow of $H$.


Indeed, by an obvious induction we can reduce to the case when $N$ is normal, let $Q$ be a $p$-Sylow of $H$. Then $Q$ is contained in a $p$-Sylow of $G$, that is, a conjugate $gPg^{-1}$. So $g^{-1}Qg\subset P\cap H$ (because $H$ is normal); by cardinality this is also a $p$-Sylow of $H$ and the lemma is proved.]


