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, $|P \cap Q|=3$.


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.

  • $\begingroup$ where did P and Q become abelian? I would understand that if the order of P and Q were prime because then they'd be cyclic and obviously abelian, but here I'm not quite seeing it because the orders of P and Q are 9. $\endgroup$ Dec 11 '13 at 1:56
  • 1
    $\begingroup$ In general, if the order of a group is the square of a prime, then this group must be abelian! $\endgroup$ Dec 11 '13 at 6:08
  • $\begingroup$ Sir, can you please help me in a different proof of same problem math.stackexchange.com/questions/1727781/… $\endgroup$
    – User
    Apr 4 '16 at 18:17
  • $\begingroup$ I see that Juanpemo already has helped you. $\endgroup$ Apr 4 '16 at 19:01

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 $|G|>5!$:


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.

  • $\begingroup$ But $P$ and $Q$ do not normalize each other, so $PQ$ does not have to be a group! Even if they are both abelian! $\endgroup$ Dec 2 '13 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.