Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product
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Let $G$ be a finite group whose order is $n$ and whose identity element is $e$.
Let $G$ be such that it has exactly $1$ Sylow $p$-subgroup for each prime divisor of $n$.
Then $G$ is isomorphic to the internal direct product of all its Sylow $p$-subgroups.
If each of the Sylow $p$-subgroups are unique, they are all normal.
As the order of each one is coprime to each of the others, their intersection is $\set e$.
|This needs considerable tedious hard slog to complete it.|
In particular: It remains to be shown that the direct product is what is is
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- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \theta$