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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.

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    I think this question showed up before. The standard counterexample is $A_4$ that has no subgroup of order $6$.2012-02-28
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    possible duplicate of [How to prove that the converse of Lagrange's theorem is not true?](http://math.stackexchange.com/questions/100933/how-to-prove-that-the-converse-of-lagranges-theorem-is-not-true)2012-02-28
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    @m. k.: Also, maybe you should read about Hall subgroups: http://en.wikipedia.org/wiki/Hall_subgroup2012-02-28
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    ego and Alex: you haven't read the question! I would guess that the answer is yes, but proving it might not be easy.2012-02-28
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    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.2012-02-28
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    BTW, I am interested in an answer too. +1 and a star! :-)2012-02-28
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    @DerekHolt You're right, I was so caught up on the title that I misread the question. I'd retract my close vote if I could. +12012-02-28
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    @DerekHolt: You are right. For m.k.: Maybe your title could be improved.2012-02-28
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    @ego: You're right, the title is misleading. I changed it.2012-02-28
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    Related: [This question](http://math.stackexchange.com/questions/33422/groups-with-order-divisible-by-d-and-no-element-of-order-d) has the same question, but with Cauchy's theorem.2012-02-28

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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 $n$.

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    Nice! Thank you for this elegant proof. Also, thanks to m.k. for a nice question.2012-02-28
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    Derek, I see this interesting post. Why is there always $c$ such that $p^a$ divides $q^c-1$?2014-11-17
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    @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 $q^c \equiv 1 \quad mod \ p^a$.2017-04-03
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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.

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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.