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In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $\langle g\rangle \oplus \ G/\langle g\rangle$ it concludes that by induction $G \cong \mathbb{Z}/d_1\mathbb{Z} \oplus \dots \mathbb{Z}/d_n\mathbb{Z}$ with specific rules on $d_i$'s!

1- So why not $G \cong \mathbb{Z}/p_1\mathbb{Z} \oplus \mathbb{Z}/p_1\mathbb{Z} \dots \mathbb{Z}/p_1\mathbb{Z} \oplus \mathbb{Z}/p_2\mathbb{Z} \dots \mathbb{Z}/p_2\mathbb{Z} \oplus \dots \dots \dots \mathbb{Z}/p_n\mathbb{Z}$?

2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)

3- And if writing them as powers of primes so why following the below steps? enter image description here

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I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $\mathbb{Z}^r$ in the decomposition.

1) We may not have necessarily have prime orders. Note $<g> \cong \mathbb{Z}/d\mathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $\mathbb{Z}/4\mathbb{Z} \not\cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$.

2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds

  1. Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.
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  • $\begingroup$ But $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$. A better example might be $\mathbb{Z}/4\mathbb{Z}$. $\endgroup$ – André 3000 Dec 2 '18 at 9:06
  • $\begingroup$ @André3000 Yes you're right. I'll edit. $\endgroup$ – Joel Pereira Dec 2 '18 at 13:30

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