# Show that a group of order $p^2q^2$ is solvable

I am trying to prove that a group of order $p^2q^2$ where $p$ and $q$ are primes is solvable, without using Burnside's theorem. Here's what I have for the moment:

• If $p = q$, then $G$ is a $p$-group and therefore it is solvable.
• If $p \neq q$, we shall look at the Sylow $p$-subgroups of $G$. We know from Sylow's theorems that $n_p \equiv 1 \pmod p$ and $n_p \mid q^2$, therefore $n_p \in \{1, q, q^2\}$.

• If $n_p = 1$, it is over, because the Sylow $p$-Subgroup $P$ is normal in $G$ of order $p^2$, and $G/P$ has order $q^2$. Thus both are solvable and $G$ is solvable.

• If $n_p = q^2$, we have $q^2(p^2-1)$ elements of order $p$ or $p^2$ in $G$, and we have $q^2$ elements left to form a unique Sylow $q$-subgroup. By the same argument as before, $G$ is solvable.
• That's where I'm in trouble. I don't know what to do with $n_p = q$. It seems to lead nowhere.

Thanks in advance for any help!

Laurent

• Welcome to math.SE! Thanks for showing your work so far! Jun 15 '14 at 16:19
• Since you reduced this to $n_p=q$, you have $n_q = p$ as well. Also $p<q$ without loss of generality. Jun 15 '14 at 16:23
• I'm not sure that I follow your argument. You seem to be assuming that distinct Sylow $p$-subgroups intersect in the identity, and that needs some justification at least, as far as I can see. This can't hold for both a Sylow $p$-subgroup and a Sylow $q$-subgroup in any case. If $P \in {\rm Syl}_{p}(G)$ has $|P| >|Q|$ for $Q \in {\rm Syl}_{p}(G),$ for example, we have $P \cap P^{g} >1$ for any $g \in G \backslash N_{G}(P),$ otherwise $|PP^{g}| > |G|.$ Jun 15 '14 at 16:50
• Yes, but what worries me is that you could have elements of order $p$ or $p^{2}$ which lie in more than one Sylow $p$-subgroup, so that $q^{2}(p^{2}-1)$ would be an overcount of the number of elements of order $p$ or $p^{2}.$ Jun 15 '14 at 18:41
• In a group whose Sylow $p$-subgroup has prime order $p,$ it is clear that distinct Sylow $p$-subgroups only have the identity in common. In general, for larger Sylow subgroups that need not be the case. In particular groups, it needs to be checked. Jun 15 '14 at 22:32

You argument works just as well with $p$ and $q$ switched, so the only time you have trouble is if both $n_p=q$ and $n_q = p$. Since $1\equiv n_p \mod p$ and $1\equiv n_q \mod q$ this puts very strong requirements on $p$ and $q$.

Hint 1:

Unless $n_p=1$, $n_p > p$.

Hint 2:

If $n_p=q$, then $q>p$. If $n_q =p$, then $p>q$. Oops.

### Fix for OP's argument:

The OP's argument is currently flawed in the case $n_p=q^2$, so this answer is only truly helpful after that flaw is fixed.

A very similar argument to the one given in this answer works. First part of your argument works, and the $p-q$ symmetry helps:

If $n_p=1$ or $n_q=1$, then the group is solvable.

Now we use the Sylow counting again to get some severe restrictions:

If $n_p \neq 1$, then $n_p \in \{q,q^2\}$ and in both cases we have $1 \equiv q^2 \mod p$. Similarly, if $n_q \neq 1$, then $1 \equiv p^2 \mod q$.

Unfortunately now we don't get an easy contradiction, but at least we only get one possibility:

Since $p$ divides $q^2-1 = (q-1)(q+1)$, we must also have $p$ divides $q-1$ or $q+1$, so $p \leq q+1$ and $q \leq p+1$, so $p-1 \leq q \leq p+1$. If $p=2$ is even, then $q$ is trapped between 1 and 3, so $q=3$. If $p$ is odd, then $p-1$ and $p+1$ are both even, so the only possibility for $q \neq p$ is $q=p-1=2$ (so $p=3$) or $q=p+1=2$ (so $p=1$, nope). Hence the only possibility is $p=2$ and $q=3$ (or vice versa).

In this case, we get:

If $p=2$ and $q=3$, then $n_q \in \{2,4\}$. Considering the permutation action of $G$ on its Sylow $q$-subgroups, we know that $n_q=2$ is impossible (Sylow normalizers are never normal) and $n_q=4$ means $G$ has a normal subgroup $K$ so that $G/K$ is isomorphic to a transitive subgroup of $S_4$ containing a non-normal Sylow 3-subgroup and having order a divisor of 36. The only such subgroup is $A_4$, so $K$ has order 3. Hence $G/K\cong A_4$ and $K \cong A_3$ are solvable, so $G$ is solvable.

• I am tempted to henceforth end all my proofs by contradiction with "oops". +1 Jun 15 '14 at 18:19
• Thank you! I think I managed to solve it with your hints! As you said, if we have $n_p = q$ and $n_q = p$, we fall on the contradiction you mentioned in the hint 2, $p > q$ and $q > p$. Thus we can conclude that this case can't happen, and therefore $G$ is solvable. Is that right? Jun 15 '14 at 18:20
• That's OK, but unfortunately your reduction to the case $n_p=q$ and $n_q=p$ is not fully justified. Jun 15 '14 at 19:07
• I gave a fix for the OP's reduction. It is probably harder than it should be. :-) Jun 15 '14 at 19:48
• I get almost everything, but the last part of the argument goes a bit beyond my current level, I think. I need a few precision: 1. as english is not my mother tongue, I don't know what an OP's reduction is. 2. $n_q= 2$ is impossible because Sylow $q$-subgroups are conjugate, and not normal? 3. I do not get the last part of the argument (since $n_q = 4$). I think I don't know enough theory for the moment, isn't there any simpler way? Jun 15 '14 at 20:04

Assuming that you know that groups of order $p^2q$, $pq$ and $p^k$ are solvable, it is enough to prove that a group of order $p^2q^2$ is not simple.

Suppose that $G$ is a simple group of order $p^2q^2$. By symmetry (and since $p$-groups are solvable) we may assume $p > q$. Steps to reach a contradiction:

1. Prove the following: if $G$ is a finite group with a subgroup $H$ of index $r$, where $r$ is the smallest prime divisor of $|G|$, then $H$ is a normal subgroup.

2. By 1. $n_p = q$ cannot happen since $G$ is simple. Therefore $n_p = q^2$.

3. If there exist distinct Sylow $p$-subgroups $P_1$ and $P_2$ such that their intersection $D = P_1 \cap P_2$ is nontrivial, then $D$ has order $p$. Now $D$ is normal in both $P_1$ and $P_2$, but not normal in all of $G$, so $N_G(D)$ has order $p^2q$. This is a contradiction by 1.

4. Therefore distinct Sylow $p$-subgroups of $G$ have pairwise trivial intersection. By 2. this means that there are $q^2(p^2-1)$ elements of order $p$ or $p^2$. But then $G$ has a normal Sylow $q$-subgroup.

• Hi! First of all, thanks for helping! I like this solution, nevertheless, if the 2., 3. and 4. points are easy to understand, the first one is quiet hard to prove. I tried to follow the steps of this post: math.stackexchange.com/questions/164244/… I just don't get the argument "since $p$ is the smallest prime that divides $|G|$, it follows that $|G/K|=p$. That would be great if you could explain, because it the only thing that I don't understand. Jun 16 '14 at 20:00
• @LaurentHayez: Since $p$ is the smallest prime divisor, $\gcd(|G|, p!) = p$. This follows since a common divisor of $|G|$ and $p!$ would have all of its prime divisors $\leq p$, hence it has to be $p$ or $1$. If it is still not clear, you could try proving it in the case $|G| = p^2q^2$. Jun 16 '14 at 21:00

I will give a different series of hints/steps to show the following: if $p >q,$ then $G$ has a non-identity normal $p$-subgroup. Let $P \in {\rm Syl}_{p}(G),$ and suppose that no non-identity subgroup of $P$ is normal in $G$ (that includes $P$ itself, of course). Notice that $q \not \equiv 1$ (mod $p$), since $1 \neq q < p.$ Hence $G$ must have $q^{2}$ Sylow $p$-subgroups, and we must have $q^{2} \equiv 1$ (mod $p$). Hence $p | q+1$ ( we can't have $p|q-1$ as $q <p$). But $q <p,$ so $q+1 \leq p,$ so we must have $q = p-1$. Now $p \neq 2$ as $p>q,$ so $q$ is even. Hence $q = 2$ and $p=3,$ as $p$ is a prime. Hence $|G| = 36$. Now $P = N_{G}(P)$ by Sylow's Theorem. Furthermore, there is non proper subgroup $M$ of $G$ which strictly contains $P.$ For otherwise we would have $[M:P] \equiv 1$ (mod $3$) and $[G:M] \equiv 1$ (mod 3), forcing $[G:P] \geq 16,$ which is not the case. Now let $g^{-1}Pg$ be another Sylow $3$-subgroup of $G.$ Then $P \cap g^{-1}Pg \neq 1$ as $|P||g^{-1}Pg| > |G|$.However $P$ and $g^{-1}Pg$ are both Abelian, so $P \cap g^{-1}Pg \lhd \langle P,g^{-1}Pg \rangle >P.$ But there is no subgroup of $G$ strictly between $P$ and $G,$ so $P \cap g^{-1}Pg \lhd G.$ (actually your (Laurent's) argument works in $G/(P \cap g^{-1}Pg)$ to show that a Sylow $2$-subgroup is normal in that quotient group.

• Thank you, I got everything except one point. It is the part "For otherwise we would have $[M:P] \equiv 1 \pmod 3$ and $[G:M] \equiv 1 \pmod 3$". I really can't see why the indexes should be equivalent to $1$ modulo $3$? Jun 16 '14 at 20:04
• Because $P$ is also the normalizer of a Sylow $3$-subgroup of $M$, we must have $[M:P] \equiv 1$ (mod $3$). Because $P$ is the normalizer of a Sylow $3$-subgroup of $G,$ we must have $[G:P ]\equiv 1$ (mod $3$). Hence we have $[G:P] = [G:M][M:P] \equiv 1$ (mod $3$) also forces $[G:M] \equiv 1$ (mod $3$). Jun 16 '14 at 22:21