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Let $G$ be a finite abelian group and $x$ an element of maximal order. Show that$\langle x \rangle$is a direct summand of $G$. Use this to obtain another proof of Theorem 2.1.

Theorem 2.1: Every finitely generated abelian group $G$ is isomorphic to a finite direct sum of cyclic groups in which the finite cyclic summands (if any) are of orders $m_1,...,m_t$, where $m_1>1$and $m_1|m_2|...|m_t$.

I want to show that $G\cong \langle x \rangle \oplus G/\langle x \rangle$. But I have difficulties in constructing the isomorphism.

The question has been answered with a method using the theorem about the structure of finite abelian groups (Finite abelian groups - direct sum of cyclic subgroup). But I don't think this exercise can be done with the theorem (or its corollaries). Thank you!

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  • $\begingroup$ This is not a duplicate. The suggested link would prove the structure theorem by using the structure theorem. $\endgroup$ – N. S. Apr 19 at 17:20
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    $\begingroup$ @Skyshie Maybe ask again the question, and state clearly: Prove this WITHOUT using the structure theorem. $\endgroup$ – N. S. Apr 19 at 17:21
  • $\begingroup$ As far as I can see the link does not use the structure theorem, but there might be more direct proofs. $\endgroup$ – Derek Holt Apr 20 at 7:20

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