Here are my notes, since I like fairly explicit answers.
In case anyone wants to see the other examples more briefly:
Elementary abelian p-groups of order pn are found TI in PSL(2,pn). Such subgroups are uniquely identified by which subspace they stabilize.
Quaternion 2-groups are found TI in the their semi-direct products with their (unique) faithful irreducible module over a prime field of odd order. The key point is that the unique element of order 2 acts fixed-point-freely on the module, and so is not contained in the intersection of any two distinct Sylow 2-subgroups.
A weaker question has a positive answer:
For every p-group P is there a finite group G such that for some g in G, P ∩ Pg = 1?
Yes. For any p-group P, take G to be the semi-direct product of P with a large enough faithful module V over a finite field of characteristic not p. Then consider the union of CV(x) as x varies over P. Since V is faithful, each centralizer is a proper subspace, and if V has large enough dimension, it cannot be written as the union of |P| proper subspaces. If v is some element of V outside that union of centralizers, then P ∩ Pv = 1.
For p = 2, the classification is due to Suzuki (1964) who discovered his infinite family of finite simple groups in this same line of investigation. Not only did he classify the possibly P, but also the possible G. Let N be the largest odd-order normal subgroup of G.
- P is cyclic and G = P ⋉ N
- P is quaternion of order 8, and either G = P ⋉ N, or G / N ≅ SL(2,3)
- P is generalized quaternion and G = P ⋉ N
- P is elementary abelian of order 2n and PSL(2,2n) ≤ G / N ≤ PΓL(2,2n)
- P is the Sylow 2-subgroup of PSU(3,2n) and PSU(3,2n) ≤ G / N ≤ PΓU(3,2n) — P is sort of a GF(2n) version of the quaternion group of order 8.
- P is the Sylow 2-subgroup of Sz(2n) and G / N = Sz(2n)
In other words, I missed two infinite families (and one fusion system). The structure of N is restricted, but not I think not classified. In some sense, one should read this list as "there exists an N such that...".
The classification for odd p appears to be a post-CFSG result, and is closely related to strongly embedded subgroups (similar to the concept Arturo mentions below: malnormal). The strongly embedded list is on page 383 - 384 of number 3 of the GLS writeup of the CFSG. The TI list is in Blau-Michler (1990), but is based on older lists I have not yet tracked down.
Other than cyclic and elementary abelian, there are a few more infinitely families, and a few sporadic exceptions.
Bibliography:
Suzuki, Michio.
"Finite groups of even order in which Sylow 2-groups are independent."
Ann. of Math. (2) 80 (1964) 58–77.
MR162841
DOI:10.2307/1970491
Blau, H. I.; Michler, G. O.
"Modular representation theory of finite groups with T.I. Sylow p-subgroups."
Trans. Amer. Math. Soc. 319 (1990), no. 2, 417–468.
MR957081
DOI:10.2307/2001249
This line seems especially mistaken: "I think, by definition of a normal subgroup, they are abelian and so this tells us that G is abelian." Certainly normal subgroups need not be abelian: for an example you can take the alternating subgroup of the symmetric group for any $n>5$.
The Sylow theorems tell you that $n_7\in \{1,5,25\}$ and that it is 1 mod 7, and so the only possibility is that it is 1.
The Sylow theorems tell you that $n_5\in \{1,7\}$ and that it is 1 mod 5, and so the only possibility is that it is 1.
Thus for both 5 and 7 you have unique (=normal for Sylow subgroups) subgroups. Let's call them $F$ and $S$ respectively. Clearly $FS$ is a subgroup of $G$ of size 175 by the reasoning you gave. (The reason that $F\cap S$ is trivial is that the intersection is a subgroup of both $F$ and $S$, so it must have order dividing both the order of $F$ and of $S$, but the greatest common divisor is 1.)
$S$ is obviously abelian, as it is cyclic (of prime order!). The question is whether or not a group of size 25 must be abelian. There are a lot of ways to see that, but the one that comes to my mind is to say that it definitely has a nontrivial center. If its center $C$ were of order $5$, then $F/C$ would be cyclic of order 5. However, by a lemma (If $G/Z$ is cyclic for a central subgroup $Z$, then $G$ is abelian) $F$ would have to be abelian.
So $G$ is a product of two abelian subgroups, and so is abelian itself.
And also, your conclusion about the two types of abelian groups of order 175 is correct. Initially you wrote that there were "two isomorphic types," but (I edited that to correct it and ) I hope that was just a slip and that you really did mean "two non-isomorphic types".
Best Answer
Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.