Before you think I have gone totally crackers:
Cauchy’s theorem says that a finite group whose order is divisible by a prime number p contains a subgroup which is cyclic of order p.
My co-authors and I have proved some similar results, of which the one referred to in the title is the following:
A finite grup whose order is divisible by 6 contains a subgroup which is either cyclic of order 6, dihedral of order 6, or isomorphic to the alternating group of degree 4 (with order 12).
When the more general theorem is proved and the paper written, I hope to elaborate on this. But my question for now is: have you seen this before?
Peter Cameron's Blog 
Perhaps relevant:
A.S.Kondrat’ev and N.A.Minigulov, Finite groups without elements of order six, Math. Notes 104 (2018), no. 5-6, 696-701.
They refer to three papers in 1977 which independently determined the simple groups with this property. This is all we need for the proof. Benjamin Sambale found the first, Rick Thomas the second, and I was unaware of the third.
Now our proof goes as follows. Let G be minimal with respect to containing both C_2 and C_3. If G is soluble there are just the three examples in our theorem. If it is not, then it has a unique maximal normal subgroup, such that the simple quotient is also minimal containing C_2 and C_3. Using the 1977 result, we see that it contains one of our three groups.
Sounds very interesting.
Is there any chance that a more general theorem holds? For instance for product of two primes? Or maybe a prime times two?
Also, it seems like this depends on the CFSG, which is disappointing. Is there any chance for a proof that does not require the CFSG?
Yiftach, your wish is my command (at least partly).
The proof for 6 doesn’t depend on CFSG, but on results about simple groups with no elements of order 6, going back to 1977 (three papers independently did this).
Meanwhile, we (really David Craven) have shown that the analogous result holds for twice a prime p if and only if p is a Fermat prime; and that it never holds for pq (for p and q odd primes) except possibly {3,5}, where there is a small detail still to be settled (but this does require CFSG).
Oh, so the proofs for the simple groups do not go by checking case by case, nice!
Also, nice result about the product of primes.
Now, I am regretting not asking for a million pounds. 🙂
In the proof for {2,3}, the only groups we have to consider are PSL(2,q), PSL(3,q) and PSU(3,q) for certain q, and all these involve PSL(2,p) where q is a power of p. For p=3 this is A_4, otherwise it contains D_6. So not much case by case.
Sorry, I should have asked, does the generalization also not depend on the CFSG?
It does depend on CFSG.