$|S_1...S_n| = |S_1|...|S_n|$(Sylow subgroups of a group $G$)

Question

Let $G$ be a finite group, and let $S_1, ..., S_n$ be its nontrivial Sylow subgroups. Assume all $S_i$ are normal in $G$.

I need to prove that $G = S_1...S_n$, that is, $\left|S_1...S_n\right| = \left|S_1\right|...\left|S_n\right|$?

Obviously, if $|G| = p^{k_1}_1...p^{k_n}_n$ is the prime factorization of the order of $G$, then ( after reordering, if needed ), $|S_i| = p^{k_i}_i$ by the definition of a $p$-Sylow subgroup.

By these considerations, $S_i \cap S_j = \left\{e \right\}$ for $i \neq j$ (a consequence of Lagrange's theorem).

But how to prove that $|S_1...S_n| = |S_1|...|S_n|$?

The proofs I have found simply say: "by order considerations".

Answer 1

Let $x\in S_i, y\in S_j,i\neq j, [x,y]=xyx^{-1}y^{-1}=x(yx^{-1}y^{-1}),$ since $S_i$ is normal, $yx^{-1}y^{-1}\in S_i$, thus $[x,y]\in S_i$,$[x,y]=(xyx^{-1})y^{-1}$ since $S_j$ is normal, $xyx^{-1}\in S_j, [x,y]\in S_i\cap S_j=\{e\}, xy=yx$.

$f:S_1\times ..\times S_n\rightarrow G, f(s_1,..,s_n)=s_1..s_n$ is well defined and injective.To see this: $s_1..s_n=1$ implies that $(s_1..s_n)^{p_2^{k_2}..p_n^{k_n}}=s_1^{p_2^{k_2}..p_n^{k_n}}=e$ This implies that $s_1=e$ $gcd(p_1,p_i)=1, i>1$. Recursively, we deduce that $s_i=e$ ans $f$ is injective, thus is an isomorphism since the source and target of $f$ have the same cardinal.

Answer 2

I found this: https://crazyproject.wordpress.com/2010/06/21/every-finite-abelian-group-is-the-direct-product-of-its-sylow-subgroups/

Actually, I had found it before posting the question, but I couldn't understand one arguement( why $S_1 \cap (S_2...S_n) = \{ e \}$ ).

Now, I get it. By induction $S_2...S_n \cong S_2 \times ... \times S_n$, so $|S_2...S_n| = |S_2 \times ... \times S_n| = |S_2|...|S_n|. DonAntonio was right, thanks to him.

So by Lagrnage's theorem $S_1 \cap (S_2...S_n) = \{ e \}$.