Length of Abelian Group
Jump to navigation
Jump to search
Theorem
Let $G$ be an abelian group whose order is $n$.
Let $n$ have the prime decomposition:
- $\ds n = \prod_{i \mathop = 1}^r p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
where $p_1 < p_2 < \cdots < p_r$ are distinct primes and $k_1, k_2, \ldots, k_r$ are positive integers.
Then the length of $G$ is given by:
- $\ds \map l G = \sum_{i \mathop = 1}^r k_i = k_1 + k_2 + \cdots + k_r$
Proof
This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. 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 {{ProofWanted}} 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 $2$: Normal and Composition Series: $\S 73 \gamma$