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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?

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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.]

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You seem to be assuming that $S_i\triangleleft S_j$ for all $j>i$, which does not automatically follow from the case $j=i+1$ as in the question. However, I suspect that the OP meant to assume the stronger version. – Neil Strickland Jun 25 '13 at 21:49
@Neil Strickland. Yes, $S_i$ is normal olny in $S_{i+1}$. In other words, $S_i$ is subnormal in $S_j$ for $j>i$. But it's O.K., the Lemma is valid in that case too. And I proved something similar too. But... – user35603 Jun 25 '13 at 22:40
But why $M$ does not depend on $n$? – user35603 Jun 25 '13 at 22:57
@user35603: As there are only finitely many possibilities for $M_n$ ($\le S_i$), at least one of them shows up infinitely times often. – j.p. Jun 26 '13 at 6:29
Thanks to all. I think you are right. – user35603 Jun 26 '13 at 11:24

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