Find a product of cyclic groups isomorphic to $(\mathbb{Z}_{12} × \mathbb{Z}_{12}) /\langle(2, 6)\rangle$.

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).

• Did you see all the questions at MSE about this type of questions? See for example here, and look at the related questions. – Dietrich Burde Jul 7 '17 at 8:05

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 $n=1,2,3$.)