Show that if $|G|=p^3q$ and $G$ has no normal Sylow subgroups, then $G \cong \mathbb S_4$

The attempt at a solution

By the Sylow theorems we have:

-$n_p \equiv 1 (p), \space n_p|q$

-$n_q \equiv 1 (q), \space n_q|p^3$

Since $G$ has no normal subgroups, one can deduce from one of the Sylow theorems that $n_p,n_q \neq 1$, so the possibilities are $n_p=q, n_q \in \{p,p^2,p^3\}$

Since $|\mathbb S_4|=2^33$ and $\mathbb S_4$ has four $3-$Sylow subgroups, I should prove that $n_q=p^2$ and, specifically, that $p=2^3,q=3$. After that, I must somehow conclude $G \cong \mathbb S_4$

I would appreciate if someone could explain me how to show these things. Answers, hints, suggestions are welcome. Thanks in advance.

  • 3
    $\begingroup$ Suppose $1+kp=q$ and $1+lq=p$ this would give $1+k(1+lq)=q\Rightarrow 1+k+klq-q=0$.. Do you see something wrong? $\endgroup$ – user87543 Sep 15 '14 at 13:54

Extended hints:

  1. You know that $n_p=q\equiv1\pmod p$, so $q>p$.
  2. Praphulla explained why you cannot have $n_q=p$ (see item 1 with roles reversed if the penny didn't drop).
  3. If you had $n_q=p^3$, then there would be $p^3(q-1)$ elements of order $q$. This would leave room for only $p^3$ other elements, which makes it impossible to ...
  4. So that leaves $n_q=p^2\equiv1\pmod q$. Thus $q\mid(p^2-1)=(p-1)(p+1)$. How does item 1 imply that $q\mid p+1$?
  5. So $p<q$ are primes such that $q\mid p+1$. In particular $q\le p+1$. How does this imply that $p=2, q=3$?
  6. So $G$ has 4 Sylow-$3$ subgroups. Let us study the conjugation action of $G$ on them. Why is the action transitive?
  7. No 3-Sylow normalizes another. Why does this imply that the conjugate action is doubly transitive?
  8. If $f:G\to S_4$ is the homomorphism coming from the above action, show that this implies that the image of $f$ is either $S_4$ or $A_4$.
  9. Why cannot it be $A_4$?
  • $\begingroup$ Probably significant streamlining in steps 6-9 is possible. This is largely "first aid". $\endgroup$ – Jyrki Lahtonen Sep 15 '14 at 14:20
  • $\begingroup$ This answer is of big help. I have some questions/doubts: in 3. suppose there are $p^3$ $q$-Sylow subgroups, so you argue that $p^3q=|G|\geq p^3(q-1)+q(p^3-1)$. But can't, for example, the $q-$ Sylow subgroups have intersection non trivial? I don't see why this can't be the case. I could follow $4.$ $q>p$ because $q=pk+1>p$ by 1. Then $q$ can't divide $p-1$, so $q$ divides $p+1$, since $q$ is prime, $q=p+1$. Then one of the primes is odd and the other is even, so it must be $p=2$ and $q=3$. I'll give it a little more thought to $6-9$. $\endgroup$ – user100106 Sep 16 '14 at 0:24
  • 1
    $\begingroup$ Remember that $q$ is a prime. Two groups of order $q$ can thus only intersect trivially by Lagrange (the order of the intersection is a factor of $q$). The same does not hold for groups of order $p^3$. The intersection of two such things can be of order $p^2$ or $p$. But having $p^3(q-1)$ elements of order $q$ means that there are only $$|G|-p^3(q-1)=p^3(q-(q-1))=p^3$$ other elements. How many subgroups of order $p^3$ can you have so that their UNION has cardinality at most $p^3$? $\endgroup$ – Jyrki Lahtonen Sep 16 '14 at 4:20
  • 1
    $\begingroup$ One subgroup, thanks for the explanation $\endgroup$ – user100106 Sep 16 '14 at 14:59
  • $\begingroup$ I know this is old but I'm stuck trying to understand why no 3-Sylow normalizes another. $\endgroup$ – Leo Lerena Apr 24 '18 at 19:04

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.