Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product
Jump to navigation
Jump to search
Theorem
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.
Proof
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 To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \theta$