If a finite group has all $p$-complements, is it always solvable? On reading the question Subgroups of Prime Power Index I immediately thought "if the group had a $p$-complement for each prime $p$ then it would be solvable".
But then I realized that the argument I had half-structured in my head was not actually correct, and I was not able to salvage it.
My basic idea was that it seems like one could apply Hall's criterion for solvability by taking suitable intersections of $p$-complements to get the desired Hall-subgroups. But on further consideration, it is not so clear why this should work.
Another idea was to do the same while looking at a minimal counterexample, but it was not clear to me why those $p$-complements should themselves satisfy the conditions (and hence, the minimality was not much use).
 A: I believe your idea is correct, it should be a matter of using the lemma that says that if the indices of the subgroups $A, B$ of $G$ are coprime, then
$$
\lvert G : A \cap B \rvert = \lvert G : A \rvert \cdot \lvert G :  B \rvert.
$$

Let us prove the lemma, let us say in the case $G$ is finite, but you only need the two indices to be finite.
We have that $\lvert G : A \rvert$ divides $\lvert G : A \cap B \rvert = \lvert G : A \rvert \cdot \lvert A : A \cap B \rvert$, and the same for $\lvert G : B \rvert$. Since the two indices are coprime, we have that their product $\lvert G : A \rvert \cdot \lvert G :  B \rvert$ divides $\lvert G : A \cap B \rvert$.
But we also have 
$$
\lvert G : A \cap B \rvert = \lvert G : A \rvert \cdot \lvert A : A \cap B \rvert = \lvert G : A \rvert \cdot \lvert A B : B \rvert \le \lvert G : A \rvert \cdot \lvert G : B \rvert.
$$
Here I have used the fact that $\lvert A B : B \rvert = \lvert A : A \cap B \rvert$, that is, under no normality assumption, there is a bijection between the cosets of $B$ contained in  $AB = \{ab :a \in A, b \in B\}$ and the cosets of $A \cap B$ in $A$. The correspondence is given by $a B \mapsto a (A \cap B)$.
This is basically as in Huppert's Endliche Gruppen I, I.2.12 and I.2.13 (1967 Ed.). 
A: Here is a full proof using the lemma provided by Andreas Caranti in his answer.
Assume that $G$ is a non-solvable finite group which has a $p$-complement for any prime $p$ and assume that $G$ is minimal with this property.
Let $p$ and $q$ be distinct primes and let $P$ and $Q$ be a $p$- and a $q$-complement in $G$ respectively. By the lemma, $P$ has a $q$-complement (namely $P\cap Q$), and since this was for arbitary $p$ and $q$, this shows that any $p$-complement in $G$ itself has $q$-complements for all primes $q$, and hence that these are all solvable by minimality of $G$.
Let $\pi$ be a set of primes. We wish to show that $G$ has a $\pi$-Hall subgroup.
Let $p$ be a prime not in $\pi$ but which divides $|G|$ (we can assume such a prime exists or a $\pi$-Hall subgroup would just be $G$). Now $G$ has a $p$-complement $P$, whose order is divisible by all primes in $\pi$, and since it is solvable, it has a $\pi$-Hall subgroup, which is thus a $\pi$-Hall subgroup of $G$.
Now $G$ is solvable by Hall's criterion (having $\pi$-Hall subgroups for all sets of primes $\pi$).
