Direct product and Sylow subgroups Let $G$ be a finite group that is equal to inner direct product of its subgroup $P$ and $Q$, where $P$ is a Sylow $p$-subgroup and $Q$ is a Sylow $q$-subgroup of $G$. If $L \le G$, prove that $L$ is equal to the inner direct product of $P \cap L$ and $Q \cap L$.
 A: The order of $G$ is $p^mq^n$ for some integers $m,n$. $L$, being a subgroup of $G$, has order $p^aq^b$ for some $a$ and $b$. $P$, being a normal Sylow subgroup, contains every subgroup of order a power of $p$, and similarly for $Q$. Thus $P\cap L$ is a normal Sylow $p$-subgroup of $L$ and $Q\cap L$ is a normal Sylow $q$-subgroup of $L$. $P\cap L$ and $Q\cap L$ intersect trivially because $P$ and $Q$ do and by cardinality must generate $L$, hence $L$ is the direct product of $P\cap L$ and $Q\cap L$.
A: Let group $G$ be a direct product of its periodic subgroups $G_1,G_2$ such that $\pi(G_1)\cap\pi(G_2)=\emptyset$ and $H\leq G$. We prove (without Sylow theorem) that $H=(H\cap G_1)\times(H\cap G_2)$. Denote $H_i=H\cap G_i$. Obviously $H_1\times H_2\leq H$. Convercely, let $a\in H$. Show that $a\in H_1\times H_2$. Since $a\in G_1\times G_2$, then $a=a_1a_2$ for some $a_i\in G_i$. We prove that $a_i\in H$. Let $|a_i|=n_i$. Since $G_1,G_2$ coprime, then $\gcd(n_1,n_2)=1$. Clearly, $a^{n_2}=a_1^{n_2}$. Since $\gcd(n_2,|a_1|)=1$, then 
$$
\langle a_1\rangle=\langle a_1^{n_2}\rangle=\langle a^{n_2}\rangle\leq H,
$$
therefore $a_1\in H$. Similarly we can prove that $a_2\in H$.
Some definitions: For $n\in\mathbb{Z}_{>0}$ $\pi(n)$ - the set of all prime divisors of $n$. Periodic element - element of finite order, periodic group - group, all of which elements are periodic. If a - periodic element of a group $G$, then $\pi(a)=\pi(|a|)$, and for periodic $G$ $\pi(G)=\cup_{g\in G}\pi(G)$. Periodic $G_1,G_2$ are coprime iff $\pi(G_1)\cap\pi(G_2)=\emptyset$.
