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According to Sylow's theorem, every finite group with order divisible by $p^k$ for some prime $p$ has a subgroup of order $p^k$. Is this the best possible result in this direction? That is, if $n$ is not a power of a prime, does there always exist a group with order divisible by $n$ that does not have a subgroup of order $n$?

EDIT: Just to clarify, I am aware that groups like this exist. The standard example seems to be $A_4$, which has order divisible by $6$ but no subgroup of order $6$. What I am looking for is a proof that a counterexample exists for any $n$ that is not a power of a prime.

Shaun
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Mikko Korhonen
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3 Answers3

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Here is a proof that the answer is yes. Suppose first that $n = p^aq^b$ with $p,q$ prime, $a,b>0$, and suppose that $p^a > q^b$. Let $c$ be minimal such that $p^a$ divides $q^c-1$ - so clearly $c > b$. Then a faithful irreducible module for the cyclic group of order $p^a$ over the field of order $q$ has dimension $c$. (You can define the action explicitly as multiplication by an element $x$ in the field of order $q^c$, where $x$ has multiplicative order $p^a$.)

Now let $G = Q \rtimes P$ be the semidirect product of an elementary abelian group $Q$ of order $q^c$ by a cyclic group $P$ of order $p^a$, using this module action. So $Q$ is the unique minimal normal subgroup of $G$. A subgroup of $G$ of order $p^aq^b$ would have a normal subgroup of order $q^b$ which would also be normal in $Q$ and hence normal in $G$, contradiction, so there is no such subgroup.

For the general case, let $n = p^aq^br$ where $r$ is coprime to $p$ and $q$. Then a direct product of $G$, as constructed above, with a cyclic group of order $r$ has no subgroup of order $n$.

Derek Holt
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  • Nice! Thank you for this elegant proof. Also, thanks to m.k. for a nice question. – William DeMeo Feb 28 '12 at 11:53
  • Derek, I see this interesting post. Why is there always $c$ such that $p^a$ divides $q^c-1$? – mathcounterexamples.net Nov 17 '14 at 10:27
  • @mathcounterexamples.net Because $\mathbb Z / p^a \mathbb Z$ is a finity group, hence $q$ has finite order in there, so a $c$ exists with $q^c \equiv 1 \quad mod \ p^a$. – ctst Apr 03 '17 at 16:55
  • Turns out there is also a reference: McCarthy, Donald. Sylow's theorem is a sharp partial converse to Lagrange's theorem, Math. Z. 113 (1970), 383–384. The construction in the paper is similar to the one given in this answer. – Mikko Korhonen Jan 28 '19 at 10:12
  • It took me some time to fill the missing details. I'm adding them for someone who had to think through them like me-
    1. The representation is irreducible because otherwise the first $0,..,c-1$ powers of the generator would be linearly dependent, which would contradict the minimiality of $c$.
    2. If we had a subgroup $H$ of size $p^a q^b$, it'd have a subgroup of size $q^b$, call it $S$, which has to lie in $Q$ (as some conjugate of it does). This means the number of $q^b$ subgroups of $H$. The intersection of $H$ with $Q$ thus has size $q^b$. Clearly $Q$ is in the centralizer of $S$.
     – Andy Sep 12 '19 at 16:13
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Good question! And, you might also look at research on the following: a CLT (Converse Lagrange Theorem) group is a finite group with the property that for every divisor of the order of the group, there is a subgroup of that order. It is known that a CLT group must be solvable and that every supersolvable group is a CLT group: however solvable groups exist, which are not CLT and CLT groups which are not supersolvable.

Nicky Hekster
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I don't know if you are familiar with Hall's theorem which gives a further partial answer to your question.

A Hall-subgroup $H$ in $G$ with regard to a set of primes $\Pi$ has the property that the index of $|G:H|$ is coprime to every element in $\Pi$.

Hall's theorem states that for solvable groups Hall-subgroups exist for every set of primes. Furthermore for a given set of primes two Hall-subgroups are conjugate.

Tile
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