3

Suppose $p,q,r$ are distinct primes. Let G be a group of order $pqr$. Does the converse of Lagrange theorem hold for this group? That is I want to know if $d$ is a divisor of $pqr$ then does there exist a subgroup of $G$ of order d?

Now by Cauchy’s theorem we know that there are subgroups of order $p,q,r$. But I don’t know if there exists subgroups of order $pq,qr,pr$.

user1382316
  • 183
  • Regarding the question in the title, there are supposedly only 4 groups of order 30, and only one of them, the symmetry group of a regular $15$-gon, isn't a nontrivial direct product. There we have the cyclic subgroup with 15 elements, and also the symmetry subgroups of a regular triangle or a pentagon. – Amateur_Algebraist Feb 04 '25 at 16:45
  • By this, if $p<q<r$, it exists for $qr$ by (1) and $pr$ by (3) (using normality). I'm not immediately sure about $pq$ though. – Joshua P. Swanson Feb 04 '25 at 16:49
  • 4
    Th3.19 of this MSc thesis claims that it's true for any squarefree order. – Amateur_Algebraist Feb 04 '25 at 16:58
  • 1
    If luckily happen that (take wlog $p<q<r$) $p\nmid q-1$, $p\nmid r-1$ and $q\nmid r-1$, then the claim is true just because $G$ is cyclic. For a "no-heavier-than-Sylow" proof, see e.g. here: https://math.stackexchange.com/a/4901551/1092170. Of course, this doesn't fit to groups of order $30=2.3.5$, because $2\mid(5-1)$. – Kan't Feb 06 '25 at 12:42

1 Answers1

7

Yes. If $p<q<r$, then there are normal subgroups of order $r$ and $rq$, and you easily find subgroups of order $rq$ and $rp$, and one of order $pq$ by the Schur-Zassenhaus Theorem.

As was pointed out in the comments, this extends to groups of order $p_1p_2 \cdots p_k$ with $p_1 < p_2 < \cdots p_k$ distinct primes. You get an ascending chain of normal subgroups of orders $p_k$, $p_kp_{k-1}$, $p_kp_{k-1}p_{k-2}$, etc. That's a standard application of Burnside's Transfer Theorem, starting with the prime $p_1$ to get a normal subgroup of index $p_1$

Derek Holt
  • 96,726