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Find (with proof) a product of cyclic groups that is isomorphic to the group $(\mathbb{Z}_{12} × \mathbb{Z}_{12}) / <(2, 6)>$.

I get the denominator has order 6, so this quotient has order 24. The solution then says that since its abelian it must be one of the following:

$\mathbb{Z_{24}}, \mathbb{Z_{12}} \times \mathbb{Z_{2}}, \mathbb{Z_{6}} \times \mathbb{Z_{2}} \times \mathbb{Z_{2}}$

Let $H = <(2,6)>$. Then $(1,0) + H$ has order 4.

Two questions here:

1) Why are only those three subgroups listed? I am confused as to when a finitely generated abelian group is decomposed as a product of its invariant factors (like the three here) vs. its elementary divisors (why isn't $\mathbb{Z_{2}} \times \mathbb{Z_{2}} \times \mathbb{Z_{2}} \times \mathbb{Z_{3}}$ an option here?) It's just not at all clear from me through looking in the literature/texts, they just seem to be presented like two alternatives even though they have different constructions (one with the divisibility requirement, one is a partition of the prime factorization).

2) Why does $(1,0) + H$ have order 4? I wrote all 6 elements out explicitly and I'm just not seeing why its not 6? (Probably a very fundamental confusion here).

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  • $\begingroup$ Did you see all the questions at MSE about this type of questions? See for example here, and look at the related questions. $\endgroup$ – Dietrich Burde Jul 7 '17 at 8:05
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1) The primary decomposition is equivalent to the invariant factor decomposition essentially because of the Chinese remainder theorem: $\mathbb{Z}_m \cong \mathbb{Z}_j \oplus \mathbb{Z}_k$ iff $m = jk$ and $j,k$ are coprime. So for instance you don't need $\mathbb{Z}_2^3 \oplus \mathbb{Z}_3$ in your list because $\mathbb{Z}_2 \oplus \mathbb{Z}_3 \cong \mathbb{Z}_6$.

2) $(1,0) + H$ has order at most $4$ because $4(1,0) = (4,0) = 2(2,6)$ belongs to $H$. (And it has order precisely $4$ because $n(1,0) \not\in H$ for $n=1,2,3$.)

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  • $\begingroup$ Thanks Alex! Very clear and to the point. Appreciate your help. $\endgroup$ – tastykakes Jul 7 '17 at 6:18
  • $\begingroup$ @tastykakes You're welcome! You are encouraged to mark satisfactory answers to your questions as "accepted" (green checkmark) on the stack exchange network as it lets others know that your problem has been solved. $\endgroup$ – Alex Provost Jul 7 '17 at 6:30

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