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classification of finitely generated abelian groups

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hbghlyj Posted 2023-3-9 16:58 |Read mode
Last edited by hbghlyj 2023-4-30 19:31Wikipedia
Suppose that $G$ is an abelian group generated by $x_1,…,x_n$. Then there are natural numbers $d_1∣d_2∣⋯∣d_n$ (if $d_i=0$ then $d_{i+1}=⋯=d_n=0$) such that $G$ is isomorphic to $ℤ/⟨d_1⟩⊕⋯⊕ℤ/⟨d_n⟩$.

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 Author| hbghlyj Posted 2023-3-9 17:01

$\operatorname{ord}(b)|\max\set{\operatorname{ord}(g)|g∈G}$ for all $b∈G$

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Lemma. If the orders $|x|,|y|$ of $x,y\in G$ are coprime, then the order of $xy$ is $|x||y|$.
Proof. If $(xy)^m = 1$, then $x^{m|y|} = 1$, so $|x|$ divides $m|y|$. Since $|x|$ and $|y|$ are coprime, this implies $|x|$ divides $m$. Similarly, $|y|$ divides $m$, so by coprimality, their product divides $m$. $_\blacksquare$

Let $a$ be an element of maximum order, and $b\neq a$. Suppose $|b|\nmid |a|$. Then there exists a prime $p$ such that $|a|=p^im$, $|b|=p^jn$, $j>i$, $p\nmid m,n$. Then $|a^{p^i}|=m$, $|b^n|=p^j$ are coprime, so $a^{p^i}b^n$ has order $p^jm>|a|$, a contradiction. Therefore, $|b|\mid |a|$ for all $b\in G$.

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 Author| hbghlyj Posted 2023-3-9 17:05
The maximum order is $d_n$. Can we use 2# to prove the classification theorem?

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