4
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Problem

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.

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    $\begingroup$ 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
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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 $A_4$?
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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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