The abelian subgroups are subgroups of the normalizer of intersection of the subgroups. Suppose $P$ and $Q$ are subgroups of order $9$ with $|P\cap Q|=3$. Why is it that $P$ and $Q$ are subgroups of the $N_G(P\cap Q)$. 

The only thing I know is that $P\cap Q$ is a subgroup of $P$ (as well as Q). Also since the order of P(same with $Q$) is $9$ (which is $3^2$) and so abelian which implies that $P\cap Q$ is normal in $P$ (also in $Q$). Is it useful? Please enlighten me.Why is it that $P$ and $Q$ are subgroups of the $N_G(P\cap Q)$.
Let $p\in P$, $q\in Q$, and $a\in P\cap Q$ be arbitrary. 

To show that $P$ and $Q$ are both subgroups of $N_G(P\cap Q)$, it suffices to show that $pap^{-1}$ and $qaq^{-1}$ are both in $P\cap Q$. Since both $P$ and $Q$ are abelian, we have that $pap^{-1} = a = qaq^{-1}$ which must be in $P\cap Q$ by assumption about $a$.

Thus, both $P$ and $Q$ are subgroups of $N_G(P\cap Q)$.



Note that this is true for any pair of abelian groups $A,B\le G$. If $a\in A$, $b\in B$, and $x\in A\cap B$, then $x = axa^{-1} = bxb^{-1}$. Thus $A,B\le N_G(A\cap B)$.
Let $p\in P$, $q\in Q$, and $a\in P\cap Q$ be arbitrary. 

To show that $P$ and $Q$ are both subgroups of $N_G(P\cap Q)$, it suffices to show that $pap^{-1}$ and $qaq^{-1}$ are both in $P\cap Q$. Since both $P$ and $Q$ are abelian, we have that $pap^{-1} = a = qaq^{-1}$ which must be in $P\cap Q$ by assumption about $a$.

Thus, both $P$ and $Q$ are subgroups of $N_G(P\cap Q)$.



Note that this is true for any pair of abelian groups $A,B\le G$. If $a\in A$, $b\in B$, and $x\in A\cap B$, then $x = axa^{-1} = bxb^{-1}$. Thus $A,B\le N_G(A\cap B)$.