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.

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    $\begingroup$ I think this question showed up before. The standard counterexample is $A_4$ that has no subgroup of order $6$. $\endgroup$
    – ego
    Feb 28, 2012 at 9:28
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    $\begingroup$ @m. k.: Also, maybe you should read about Hall subgroups: en.wikipedia.org/wiki/Hall_subgroup $\endgroup$ Feb 28, 2012 at 9:52
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    $\begingroup$ 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, 2012 at 9:58
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    $\begingroup$ 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, 2012 at 10:01
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    $\begingroup$ BTW, I am interested in an answer too. +1 and a star! :-) $\endgroup$
    – user21436
    Feb 28, 2012 at 10:02

3 Answers 3


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, 2012 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, 2014 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, 2017 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, 2019 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, 2019 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.


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