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$|G|=pqr^2$ 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.

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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. –  Jack Schmidt Mar 27 '11 at 16:03

3 Answers 3

up vote 5 down vote accepted

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.

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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? –  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.

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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. –  Jack Schmidt Mar 27 '11 at 15:59
    
oops, sorry I thought we had $r>pq$ –  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.

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