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Opening any book of modern algebra, one can find that the proof of the theorem in title involves may small lemmas. I was trying to find if one can present the proof with minimum requirements, so that it is possible to present it in a lecture of one hour. Here is my way of doing it; I wanted to know if something is missing in my key-steps.

(1) Let $G$ be a finitely generated abelian group, say generated by $k$ elements.

(2) Take direct sum of $k$ copies of $\mathbb{Z}$; write it $\mathbb{Z}^k$.

(3) Lemma 1. There exists $\phi:\mathbb{Z}^k\rightarrow G$, an epimorphism with kernel $M$. So $\mathbb{Z}^k/M\cong G$.

(3) (Lemma 2.) There exists a basis $\{x_1,\cdots, x_k\}$ of $\mathbb{Z}^k$ and integers $d_1,\cdots, d_r$ ($r\leq k$) such that $M$ has a basis $\{d_1x_1, \cdots, d_rx_r\}$ and $d_i|d_{i+1}$.

(4) Then $G=\mathbb{Z}^k/M = \mathbb{Z}/d_1 \oplus \cdots \oplus \mathbb{Z}/d_r \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}$, where $\mathbb{Z}$ appears $k-r$ times.

Question 1.: Am I missing some steps in argument (except proof of Lemmas)?

Question 2. Can you suggest a reference in which an elementary proof of structure of (f.g.) abelian groups (i.e. proof with minimum lemmas) is presented?

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  • $\begingroup$ One hour is definitely a bit short if you think of proving lemma 2 and (4) without any assumptions $\endgroup$ – JJR May 22 '17 at 10:34
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For Lemma 1, let $\{g_1,\ldots,g_k\}$ be a generating set of $G$. Then each element of $G$ has the form $g_1^{m_1}\cdots g_k^{m_k}$ for some integers $m_i$. Then the mapping $\phi:\mathbb{Z}^k\rightarrow G$ given by $e_i\mapsto g_i$ is a group epimorphism, where $e_i$ denotes the $i$-th unit vector. By the homomorphism theorem, the isomorphy $\mathbb{Z}^k/M\equiv G$ follows.

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  • $\begingroup$ I am not asking for proofs of lemmas; did you see question properly. $\endgroup$ – Beginner May 22 '17 at 9:00

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