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

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.

• en.wikipedia.org/wiki/Nilpotent_group – Martin Brandenburg Jun 28 '14 at 23:22
• The title could be interpreted as asking a different question from the body. It's not true that a finite group is the direct product of its Sylow subgroups; this property characterizes nilpotent groups. – Qiaochu Yuan Jun 28 '14 at 23:46
• Finite solvable groups are characterized by the property that one can choose the $P_i$ such that $P_i P_j$ are subgroups. Finite nilpotent groups by the property that $[P_i,P_j]=1$. – Jack Schmidt Jun 29 '14 at 1:29
• @MartinBrandenburg I'm not talking about a direct product, just the element-wise product. – Avi Steiner Jun 29 '14 at 1:59
• @AviSteiner: You can pick $g_i$ to be certain powers of $g$. You can just apply the chinese remainder theorem to the cyclic group generated by $g$, I think – spin Jun 29 '14 at 18:09

## 2 Answers

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.

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.

• This actually completely answers my question, even without the GAP computation, since the "theorem" (which apparently I mis-remembered) applied to $A_6$ says that $A_6$ is the element-wise product of any triple Sylow $p$-subgroups (with distinct $p$'s). – Avi Steiner Jun 29 '14 at 16:39

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.

• This doesn't work. There's no reason I can see a priori that $P_1\cdots P_{n-1}$ is necessarily a group or even that it's intersection with $P_n$ is trivial. If it's not a group, then we can't apply the theorem you state. – Avi Steiner Jun 29 '14 at 1:58
• @AviSteiner Oh crap, I see what you mean. Since it's okay that $P_1\cdots P_n$ isn't a subgroup, I forgot that $P_1\cdots P_{n-1}$ had to be a subgroup for the induction hypothesis. I will leave this answer up as a cautionary tale in case other readers make the same mistake. – Alexander Gruber Jun 29 '14 at 21:50