Intersection of two finite abelian subgroups I have a question concerning a proof that a group of order 144 is not simple.
Given two Sylow 3-subgroups, $P$ and $Q$, we know that $P$ and $Q$ are both abelian as they are of order $p^2$. Let $M=N_G(P \cap Q)$. Then $P \cap Q$ is a normal subgroup and therefore normal in both $P$ and $Q$. This is where I am running in to a problem, the proof goes on to conclude that since $P$ and $Q$ are subset of $M$, then $PQ \subseteq M$. I am not sure how to prove this to myself. 
Also, $|P \cap Q|=3$. 
 A: Let $G$ be a group of order 144 and assume that $G$ is simple. We will argue through an analysis of the Sylow 3-subgroups of $G$ to arrive at a contradiction. On the way we will use a result maybe less known.
Lemma Let $G$ be a group and $p$ a prime dividing the order of $G$. Assume that for every pair $P, Q \in Syl_p(G)$, $P=Q$ or $P \cap Q = 1$. Then $n_p=|Syl_p(G)| \equiv 1$ mod $|P|$.Note that this generalizes the well-known result that $n_p \equiv 1$ mod $p$. I will not prove the lemma, but to sketch it: fix a $P \in Syl_p(G)$ and let this group act on the set of all Sylow $p$-subgroups by conjugation. The orbit of $P$ itself is $\{P\}$ and if $Q \neq P$, the orbit of $Q$ has length $|P|$.Let's get on with the analysis. Since $G$ is simple there are at least two different $P, Q \in Syl_3(G)$. Put $D = P \cap Q$. Of course $D$ is a proper subgroup of $G$. We are going to show that in fact $D=1$. In that case we can apply the lemma, $n_3 \equiv 1$ mod 9 and together with the fact that $n_3 \in \{1,2,4,8,16\}$, this yields $n_3 =1$, which means that the Sylow 3-subgroup is normal, against our assumption $G$ being simple.Now suppose $D \neq 1$. Since $P \neq Q$, $D$ is a proper subgroup of $P$, but $|P|=9$, so $|D|=3$. Put $H=N_G(D)$. Indeed $P,Q \subset H$, because $P$ and $Q$ are abelian. Note that since $P$ and $Q$ are different subgroups, $P\neq H$ and index$[H:P] \neq 2$, so index$[H:P] \geq 4$. On the other hand, index$[G:H]$ cannot be $2$, otherwise $H$ would be normal in the simple group $G$, so it must be a divisor of $16$ and be at least $4$. All this can only be the case if $|H|=36$, or equivalently, index $[G:H]=4$. Of course core$_G(H)=1$ and this means that $G/$core$_G(H)=G$ can be embedded in $S_4$, which is absurd since $144 \nmid 24$. This is the final contradiction and hence $G$ cannot be simple.
A: For the record, a slightly different argument to prove that $3$-Sylows intersect trivially, needed for Nicky's answer. This works for many other orders.
Let $P$ and $Q$ be two $3$-Sylow subgroups and suppose that $C=P\cap Q$ is non-trivial, so that it is in fact cyclic of order $3$. As $P$ and $Q$ are abelian, we have that $P\cup Q\subseteq N_G(C)$ and therefore the normalizer $N_G(C)$ contains the set $\{ab:a\in P,b\in Q\}$ which has $|P|\cdot|Q|/|P\cap Q|=27$ elements. It follows from this that $$[G:N_G(C)]\leq\frac{144}{27}<6.$$ This tells us that set set of subgroups conjugate to $C$ in $G$ has at most $5$ elements, and since $C$ is not normal in $G$, at least $2$. Since $G$ acts transitively on this set, we thus have a non-constant morphism $G\to S_m$ with $2\leq m\leq 5$. This is absurd, since $|G|>5!$:
A: As $P$ and $Q$ are both abelian, each normalizes $P\cap Q$, so by definition is in $M$.  Let me add that it may not be true that $PQ$ is a group.  In general, if $P$ and $Q$ are subgroups of $M$, then $\langle P, Q\rangle$ is a subgroup of $M$.  However, in this case $\langle P, Q\rangle= PQ$ because $P$ and $Q$ are abelian.
