Subgroups of groups of order 36 
*

*Is there any group of order 36 with no subgroup of order 6?

*Is there any group of order $p^2q^2$ with no subgroup of order $pq$?

*Is there any group of order $p^{2m}q^{2m}$ with no subgroup of order $p^mq^m$?

*Is there any group of order $p^{2m}q^{2n}$ with no subgroup of order $p^mq^n$?
($p,q$ are distinct primes)
 A: For question 2, I believe the answer is yes, which also settles question $3$ and $4.$  Take $p = 149$ and $q = 5.$ Then a cyclic group of order $25$ acts irreducibly and faithfully on an elementary Abelian $p$-group of order $p^{2}.$  Furthermore, in the resulting semidirect product, each element of order $5$ acts irreducibly on the normal Sylow $p$-subgroup, so there is no subgroup of order $745 = pq.$ 
The general strategy is to find primes odd primes $p,q$ with $p \equiv -1$ mod $q^{2},$ and then a similar construction works.
A: This is not really a complete answer (yet). But since it's an interesting question and since I cannot post comments due to forum restrictions, I'll hint you to an article I found: http://matwbn.icm.edu.pl/ksiazki/fm/fm92/fm9211.pdf (if your browser doesn't display anything, try downloading).
On page 2 you'll find the following theorem:

My knowledge of algebra is too rusty to really understand this at first glance. But a CLT-group is a group, for which the converse Lagrange Theorem holds true, so for every divisor of the group order exists a subgroup. If this paper is right, this would mean every group of order 36 is a non-CLT group (case II), so there exist at least one divisor of 36, which has no corresponding subgroup.
Since Cauchy's Theorem and Sylow Theorems imply the existence of subgroups of order 2,3,4 and 9, the divisors left are: 6,12 and 18. This doesn't directly answer your question, but the isomorphism should help finding an answer. I'll update, as soon as I find out something more precise...
