"Let $|G|=p_1^{e_1}\cdot \cdot \cdot p_t^{e_t}$ and let $G_{p_i}$ be the Sylow $p_i$-subgroup of $G$. The subgroup $S$ generated by all of the Sylow subgroups is $G$, for $p_i^{t_i}|~|S|$ for all $i$."
The above statement is made in Rotman's Advanced Modern Algebra on page 273. I don't understand why it is true. Could someone show me what $S$ looks like? I think, $S=\{a_1^{e_1}\cdot \cdot \cdot a_n^{e_n} | \forall i, a_i \in \bigcup_{i=1}^{t}G_{p_i},e_i \in \mathbb{Z}, a_i \neq a_{i+1}, n \in \mathbb{N} \}$, but I am not sure.
Thanks.
Here it is in words in case it helps.
Simply the order of a subgroup divides the order of the group. Each Sylow subgroup of $G$ is a subgroup of $S$ by the definition of $S$, so $S$ must have order divisible by $p_i^{e_i}$. So the order of $S$ is divisible by the product of the $p_i^{e_i}$ (since these are coprime), and this product is the order of $G$.
We find that $S$ is a subgroup of $G$ with order at least as big as the order of $G$ so $S=G$ is the only possibility.