In discussing old qualifying exam problems with my classmates, I've discovered an implicit assumption I've been making about Sylow's theorem that I've never seen a proof or counter example to. It's become quite a debate, now, so we would be quite grateful for any help you can provide.
The short question is this: Taking into account possible isomorphism structures of the Sylow p-subgroups themselves, does Sylow's theorem give an upper bound for the possible number of isomorphism classes for groups of a given order?
A couple of examples to illustrate what I'm asking:
Example 1: Suppose you're considering the groups of order $2013 = 61*11*3$. Sylow's Theorem tells us that the subgroups of order 61 and 11 are normal, and that there are either 61 or 1 subgroups of order 3. Since, for each of 3, 61, and 11 there is only one isomorphism class of groups of that order (namely, the cyclic groups), can I now state that there are at most two isomorphism classes of groups of order 2013. In particular, if I now demonstrate two non-isomorphic groups of order 2013, can I conclude (with no additional arguments) that there are exactly two isomorphism classes?
Example 2: Suppose now you are considering groups of order $12 = 2^2*3$. Sylow's Theorem tells us that there are either 1 or 3 subgroups of order 4, and 1 or 4 subgroups of order 3. Since there are two isomorphism classes of groups of order 4 and one isomorphism class of groups of order 3, can I now conclude (with no additional arguments) that there are at most 8 isomorphism classes of groups of order 12?
It should be noted that we have solved both of these problems using alternative arguments, so the solutions to them are not my question. My question is about the argument I would like to make about Sylow's Theorem. If anyone has a reference, proof, or counterexample to this question, or additional assumptions that are needed, that would be extremely helpful!
If I have understood your question correctly, then $273 = 3 \times 7 \times 13$ is a counterexample.
All Sylow subgroups are cyclic, there are unique Sylow $p$-subgroups for $p=7$ and $13$, and $1$, $7$, $13$ or $91$ Sylow $3$-subgroups.
So according to your conjecture, there should be at most four isomorphism types of groups of this order, but in fact there are $5$, with two isomorphism classes of groups with $91$ Sylow $3$-subgroups.