I'm trying to prove:
Let $\lbrace P_i: i \in I \rbrace$ be a set of Sylow subgroups of a finite group G, one for each prime divisor of $|G|.$ Then $\langle \bigcup_{i \in I} P_i \rangle = G$.
(From J.J. Rotman's Introduction to the Theory of Groups, problem 4.10)
If G has only one prime divisor this is obvious; if it has two, I noticed that (and it's not hard to prove) $P_1 \cap P_2 = \lbrace 1 \rbrace$ and $\langle P_1 \cup P_2 \rangle = \langle P_1 P_2 \rangle$ (where $P_1 P_2$ is the subgroup product), which means $|\langle P_1 \cup P_2 \rangle| \geq |P_1 P_2| = |P_1| |P_2|/|P_1 \cap P_2| = |P_1| |P_2|/1 = |G|$, so the generated subgroup must be the entire group.
Now, I would like to somehow extend this result to any finite-order G by, for example, inducting on the number of prime factors of G or the number of subgroups of a G whose order has an arbitrary number of prime factors having the subgroup generated by union counted. It would be enough if I could show $|\langle \bigcup_{i=1}^n P_i \rangle| \geq \prod_{i=1}^n |P_i|$, for example, but a similar argument to the above won't work even with just three Sylow subgroups unless I can somehow exclude the possibility that some pair $x \in P_1$, $y \in P_2$ have $|xy|$ divisible by the prime associated with $P_3$.
It seems like wherever I look for a good way to count the elements of $\langle \bigcup_{i=1}^n P_i \rangle$ I run into something that would be easy in a "nice" case like if G were abelian or if some particular subgroup involved were normal, but I'm reasonably sure I can't assume anything like that is true, nor work my way up from a "nice" case to the general one.
Am I missing something simple to finish this argument, or should I be looking for a different way to count the elements of the generated group?
(P.S., there is another question on this site regarding this problem, but the answers to the one I saw stopped off right where I'm stuck.)
Let $H=\langle \bigcup_i P_i\rangle$. From $P_i\le H$ we conclude $|P_i|\mid |H|$. As the $|P_i|$ are pairwise coprime we have $|G|$ is the least common multiple of all $|P_i|$, hence $|H|$ is a multiple of $|G|$, which is only possible if $G=H$.
Put differently, assume $H\ne G$ and find a prime $p$ with $p\mid[G:H]$. Let $P$ be a $p$-Sylow subgroup of $G$ and $|P|=p^r$, say. Then $r\ge 1$ and $p^{r+1}\nmid |G|$, hence $p^r\nmid H$ and $P\not\le H$.