$|G|=pqr^2$ where $3\leq p<q<r$ prime. Show that if $r>\frac{1}{2}(pq-1)$ then $G$ is solvable.

I took $H\leq G$ $r$-sylow subgroup of $G$, there is a theorem claiming that there exists a homomorphism from $G$ to $S_{pq}$ ($pq$ is the index of $H$).

If the homomorphism is injective then $G$ is a subgroup of $S_{pq}$, that means $pqr^2|(pq)!$ so $r^2|(pq-1)!$ but we know $r^2>2r>pq-1$ from what follows that the homomorphism can't be injective, that means there is a non trivial kernel which is a maximal subgroup of $H$ and a normal subgroup of $G$, it can be of order $r$ or $r^2$, lets name it K.

If $|K|=r^2$ then the sequence $\{e\}\triangleleft K\triangleleft G$ has a factor of order $r^2$ which is abelian as a square of a prime and a factor of order $pq$.

The other option is $|K|=r$, then the factors are of order: $r$ and $pqr$.

I am a bit stuck from here.

  • 1
    $\begingroup$ I don't know if it matters, but if |G| = pqrr with 2 ≤ p < q < r, then G is solvable. Squaring the largest prime factor doesn't help to get nonsolvable groups. You need to square the smaller ones. $\endgroup$ – Jack Schmidt Mar 27 '11 at 16:03

Groups of order $r$ are solvable. Groups of order $pqr$ are solvable. So your group is an extension of two solvable groups, so solvable. In a group of order $pqr$, the Sylow $r$-subgroup is normal, and in the quotient of order $pq$, the Sylow $q$-subgroup is normal.

  • $\begingroup$ did you mention a group of order $pqr$ is solvable, once they are all prime and regardless of duplicity? May I ask how to get that? $\endgroup$ – Honghao Aug 30 '12 at 16:45

You are left to show that a group of order $pqr$ is solvable, and this can be done exactly the same way you began the $pqr^2$ case.

  • $\begingroup$ When p=3, q=5, r=11, Sym(pq) does not have order prime to r. The order just is not divisible by r*r. $\endgroup$ – Jack Schmidt Mar 27 '11 at 15:59
  • $\begingroup$ oops, sorry I thought we had $r>pq$ $\endgroup$ – Plop Mar 27 '11 at 16:00

There are two cases to consider: either the Sylow $r$-group is maximal, or it is normal in $G$. Both cases lead to a solvable group. No conditions on $r$, further than $p < q< r$, are necessary.


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.