Let $\lbrace P_i : i\in I \rbrace $ be a set of Sylow subgroups of a finite group G, one for each prime divisor $p_i$ of $|G|$. Show that $G$ is generated by $\bigcup P_i$.
from Rotman "An Introduction to the theory of groups" pag.81 n.4.10
my attempt was to demonstrate that $\langle P_1,\dots,P_j \rangle \cap P_{j+1} = 1_G$ but i know how,to prove this only in the ( trivial ) abelian case (in that case is easy to demonstrate the entire exercise)
I tried by contradiction: let $b \in \langle P_1,\dots,P_j \rangle \cap P_{j+1} $ so the order of $b$ is a power of $p_{j+1}$ but in $ \langle P_1, \dots,P_j \rangle $ there are lots of elements and i don't found the way to get an absurd.
Any hint will be appreciate, even different and more elegant solutions 🙂
Best Answer
Hint: If a finite group $G$ has a subgroup $H$ then the order of $G$ is divisible by the order of $H$ (Lagrange).
Consider the group generated by all the $P_i$ - what can you say about its order?