Is a finite group always a element-wise product of Sylow subgroups?

Question

Let $G $ be a finite group, and let $ p_1, \ldots, p_n $ be the distinct primes dividing $|G|$. For each $i $, let $ P_i $ be a Sylow $ p_i $-subgroup of $ G $. I seem to recall a theorem saying $ G=P_1\cdots P_n $. Is this true? It's immediate via a simple counting argument when $ n=2$, but I haven't been able to prove/disprove the general case.

Answer 1

I believe there is no such set of Sylow $p$-subgroups for the simple group of order 360. I'll try to write up a non-computational proof, but I just asked GAP to try all triples of Sylow $p$-subgroups.

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Calculations so far: There are two $G$-classes of $HK$ for $H$ a Sylow $3$-subgroup and $K$ a Sylow $5$-subgroup (this can be done by hand).

As Avi worried, $P \cap HK$ may have size larger than 1 (only size 1 or 2 are possible in this group). Even if $HK$ is not a subgroup, $|P \cap HK| > 1$ means $PHK \neq G$ by cardinality considerations. As an example,$P=\langle (3,4)(5,6), (3,6)(4,5), (1,2)(3,4) \rangle$, $H=\langle (1,2,3), (4,5,6) \rangle$, $K=\langle (2,5,3,6,4) \rangle$.

More precisely, we have $xhk = x' h'k'$ iff $hk (h'k')^{-1} = x^{-1} x' \in P \cap (HK)(HK)^{-1} = P \cap HKH$. Hence we just need to compute all possible $HKH$ and then their intersections with a Sylow $2$-subgroup $P$.

I don't see a smooth way to do this, however, again there are only two $G$-classes of $HKH$, and for each class we either have $|P \cap HKH|=3$ or $|P \cap HKH|=4$ (if you care: depending on the Sylow $2$-subgroup $P$, but not on the Sylow $3$-subgroup $H$ or the Sylow $5$-subgroup $K$, which is kind of weird). In particular, $|PHK|<360$ (and if you care, $|PHK| \in \{ 240, 280 \}$).

I don't see any non-computational way to handle $P\cap HKH$, but I've now triple checked the calculations, so I'm pretty confident.

Answer 2

Given subgroups $H,K\leq G$, the cardinality of $HK$ (whether it's a subgroup or not) is $$\frac{|H||K|}{|H\cap K|}.$$ Therefore, by induction, $|P_1\cdots P_n|=|G|$.

EDIT: This is wrong. See below.