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.

  • 3
    $\begingroup$ I think this question showed up before. The standard counterexample is $A_4$ that has no subgroup of order $6$. $\endgroup$
    – ego
    Feb 28 '12 at 9:28
  • 2
    $\begingroup$ @m. k.: Also, maybe you should read about Hall subgroups: en.wikipedia.org/wiki/Hall_subgroup $\endgroup$ Feb 28 '12 at 9:52
  • 9
    $\begingroup$ ego and Alex: you haven't read the question! I would guess that the answer is yes, but proving it might not be easy. $\endgroup$
    – Derek Holt
    Feb 28 '12 at 9:58
  • 3
    $\begingroup$ I think this is not a duplicate. OP is asking about existence of a group for each $n$, such that $n$ divides the order of the group but, there is no subgroup of order $n$. Well, I am sure a general infinite family is not possible but, there might be a reasonable answer. Further, what Dennis suggests will answer the converse of the question. Hall subgroups in Solvable groups. $\endgroup$
    – user21436
    Feb 28 '12 at 10:01
  • 2
    $\begingroup$ BTW, I am interested in an answer too. +1 and a star! :-) $\endgroup$
    – user21436
    Feb 28 '12 at 10:02

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 $n$.

  • $\begingroup$ Nice! Thank you for this elegant proof. Also, thanks to m.k. for a nice question. $\endgroup$ Feb 28 '12 at 11:53
  • $\begingroup$ Derek, I see this interesting post. Why is there always $c$ such that $p^a$ divides $q^c-1$? $\endgroup$ Nov 17 '14 at 10:27
  • $\begingroup$ @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 $q^c \equiv 1 \quad mod \ p^a$. $\endgroup$
    – ctst
    Apr 3 '17 at 16:55
  • $\begingroup$ 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. $\endgroup$ Jan 28 '19 at 10:12
  • $\begingroup$ 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 $S$. $\endgroup$
    – Andy
    Sep 12 '19 at 16:13

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.


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.


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.