My question is the like the title
'($\forall$ finite group, G) $\simeq$ ($\Pi$ the Sylow $p$ subgroups of the G)'
I want to know this statement holds or not.(I'm not sure my proof is correct)
My proof) I'm not sure my trial is correct or not.
When the $\vert G\vert = p^n$, it is trivial that G is isomorphic with sylow p subgroup itself.
Let $\vert G \vert$ = $p_{1}^n p_{2}^m k$ for prime $p_1,p_2$ and $k\neq 1$ with $(p_1, k) = (p_2,k) =1$
Then, $\exists P_1$ and $P_2$ $s.t.$ $\vert P_1 \vert = p_1^n$ and $\vert P_2 \vert = p_2^m$ respectively.
(By sylow First thm, $P_1$ and $P_2$ are sylow subgroups.)
Also $P_1 \cap P_2 = \{ e \}$ ($e$ means identity of the $G$)
And clearly, $P_1, P_2 \leq G$ which means subset(subgroups)
Hence, $\vert P_1P_2 \vert$ = ${\vert P_1 \vert \vert P_2 \vert} \over \vert P_1 \cap P_2 \vert$ = $\vert P_1 \vert \vert P_2 \vert$
Therefore $k$ can be prime decomposed by the other primes, $p_i(\neq p_1, p_2)$
Same process Like the above, We can conclude
$\vert G \vert$ = $\Pi _1 ^n \vert P_i \vert \Rightarrow G \simeq \Pi _1 ^n \ P_i$ (Since each $P_i$ are subsets of the G )
Best Answer
Your statement is incorrect. The correct statement is:
A finite group is isomorphic to the direct product of it's Sylow subgroups if and only if all of the group's Sylow subgroups are normal.
See for instance Theorem 3.3 of this blurb of Keith Conrad.