How many different abelian groups exist, which include subgroup $ \mathbb Z/12 \mathbb Z$ How many different abelian groups exist which include subgroup $\mathbb Z/12 \mathbb Z$ and factor by this subgroup is also isomorphic to $\mathbb Z/12 \mathbb Z$?
My suggestion is that we should have group of 144 elements (because of 12 elements in subgroup and 12 classes).
So, group of $144 $ elements can be decomposed into these 4 different direct sums:
$\mathbb Z/144 \mathbb Z$
$\mathbb Z_{12} \times \mathbb Z_{12}$
$\mathbb Z_{3} \times \mathbb Z_{4} \times \mathbb Z_{12}$
$\mathbb Z_{3} \times \mathbb Z_{4} \times \mathbb Z_{3} \times \mathbb Z_{4}$
which includes subgroups isomorphic to  $\mathbb Z/12 \mathbb Z$
And it means that the answer is "4".
Am I right?
 A: A consequence of the Chinese Remainder theorem is that, if $n = ab$, then
$$\mathbb{Z}_{n} \cong \mathbb{Z}_a \times \mathbb{Z}_b \iff \gcd(a,b) = 1.$$
The prime decomposition of $144 = 2^4 \cdot 3^2$, so we can list all Abelian groups of order $144$ (up to isomorphism) as follows:


*

*$\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_3 \times \mathbb{Z}_3$

*$\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_9$

*$\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_3 \times \mathbb{Z}_3$

*$\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_9$

*$\mathbb{Z}_{2} \times \mathbb{Z}_{8} \times \mathbb{Z}_3 \times \mathbb{Z}_3$

*$\mathbb{Z}_{2} \times \mathbb{Z}_{8} \times \mathbb{Z}_9$

*$\mathbb{Z}_{4} \times \mathbb{Z}_{4} \times \mathbb{Z}_3 \times \mathbb{Z}_3$

*$\mathbb{Z}_{4} \times \mathbb{Z}_{4} \times \mathbb{Z}_9$

*$\mathbb{Z}_{16} \times \mathbb{Z}_3 \times \mathbb{Z}_3$

*$\mathbb{Z}_{16} \times \mathbb{Z}_9$


We now have to determine which of these have a subgroup isomorphic to $\mathbb{Z}_{12} \cong \mathbb{Z}_3 \times \mathbb{Z}_4$. This is the case if and only if you have a component $\mathbb{Z}_n$ with $4|n$ and a componenet $\mathbb{Z}_m$ with $3|m$, hence:


*

*Nope.

*Nope.

*Yes, take the subgroup generated by $(0,0,1,1,0)$

*Yes, take the subgroup generated by $(0,0,1,3)$.

*Yes, take the subgroup generated by $(0,2,1,0)$.

*Yes, take the subgroup generated by $(0,2,3)$.

*Yes, take the subgroup generated by $(1,0,1,0)$.

*Yes, take the subgroup generated by $(1,0,3)$.

*Yes, take the subgroup generated by $(4,1,0)$.

*Yes, take the subgroup generated by $(4,3)$.


Finally, the quotient must also be isomorphic to $\mathbb{Z}_{12} \cong \mathbb{Z}_3 \times \mathbb{Z}_4$. Note that $\mathbb{Z}_8 / \mathbb{Z}_4 \cong \mathbb{Z}_2$; $\mathbb{Z}_{16} / \mathbb{Z}_4 \cong \mathbb{Z}_4$ and $\mathbb{Z}_9 / \mathbb{Z}_3 \cong \mathbb{Z}_3$, hence:


*Quotient is $\mathbb{Z}_2 \times \mathbb{Z}_2 \times (\mathbb{Z}_4/\mathbb{Z}_4) \times (\mathbb{Z}_3/\mathbb{Z}_3) \times \mathbb{Z}_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$

*Quotient is $\mathbb{Z}_2 \times \mathbb{Z}_2 \times (\mathbb{Z}_4/\mathbb{Z}_4) \times (\mathbb{Z}_9/\mathbb{Z}_3) \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$

*Quotient is $\mathbb{Z}_2 \times (\mathbb{Z}_8/\mathbb{Z}_4)  \times (\mathbb{Z}_3/\mathbb{Z}_3) \times \mathbb{Z}_3\cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$

*Quotient is $\mathbb{Z}_2 \times (\mathbb{Z}_8/\mathbb{Z}_4)  \times (\mathbb{Z}_9/\mathbb{Z}_3) \cong\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$

*Quotient is $(\mathbb{Z}_4/\mathbb{Z}_4) \times \mathbb{Z}_4 \times (\mathbb{Z}_3/\mathbb{Z}_3) \times \mathbb{Z}_3\cong \mathbb{Z}_4 \times \mathbb{Z}_3$

*Quotient is $(\mathbb{Z}_4/\mathbb{Z}_4) \times \mathbb{Z}_4 \times (\mathbb{Z}_9/\mathbb{Z}_3) \cong\mathbb{Z}_4 \times \mathbb{Z}_3$

*Quotient is $(\mathbb{Z}_{16}/\mathbb{Z}_4)\times (\mathbb{Z}_3/\mathbb{Z}_3)  \times \mathbb{Z}_3 \cong\mathbb{Z}_4 \times \mathbb{Z}_3$

*Quotient is $(\mathbb{Z}_{16}/\mathbb{Z}_4)\times (\mathbb{Z}_9/\mathbb{Z}_3)  \cong\mathbb{Z}_4 \times \mathbb{Z}_3$


So, yes, the answer is indeed $4$ - but the way to get there is quite different from what you suggested!
A: Observe that $\;\Bbb Z_3\times\Bbb Z_4\cong\Bbb Z_{12}\;$ , so your second, third and fourth groups are the same (up to isomorphism).
Can you see how to fix your answer now?
