Express $H$ as a product of cyclic groups 
Let  $\mathbb{Z}_n$ denote the additive cyclic group of order $n$. Let $G$ be the group $\mathbb{Z}_4 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_{11}$ and let $H$ be the subgroup $\{6g \mid g \in G\}$.
  
  
*
  
*Express the groups $H$ and $G/H$ as direct sums of cyclic groups.
  
*Is $G$ isomorphic to $H \oplus G/H$?
  

So, given $(a,b,c)\in G$, $a$ can be $0,1,2,3$, $b$ can be $0,1,2,3,4,5$ and $c$ can be $0,1,...,10$. If $6(a,b,c)=(p,q,r)$ then $p$ can be $0,2$, $q$ can be $0$ and $r$ can be $0,1,...,10$. Hence, $H=\mathbb{Z}_2\oplus \mathbb{Z}_{11}$. Then $G/H=\mathbb{Z}_3\oplus \mathbb{Z}_4$. I just wanted to make sure whether my answers and justifications are correct. Appreciate if point out any errors.
 A: Let $(0,0,z)\in G$; then, since $6\cdot2\equiv1\pmod{11}$, we have
$$
(0,0,z)=6(0,0,2z)
$$
Similarly,
$$
(2,0,z)=6(1,0,2z)
$$
so
$$
H=\{(x,0,z):x\in 2\mathbb{Z}_4,z\in\mathbb{Z}_{11}\}
$$
Now it's clear that $H\cong\mathbb{Z}_2\oplus\mathbb{Z}_{11}$.
Also
$$
G/H\cong\mathbb{Z}_4/2\mathbb{Z}_4\oplus\mathbb{Z}_6
\cong\mathbb{Z}_2\oplus\mathbb{Z}_{6}
\cong\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{3}
$$
A: Your justification for $H$ is good, but I think you're wrong about $G/H$. Why did you choose that instead of $\mathbb{Z}_2\oplus \mathbb{Z}_6$? The $q$ coordinate is fixed, and so the middle coordinate isn't restricted why you mod out by $H$.
$\mathbb{Z}_2\oplus\mathbb{Z}_6$ Contains multiple copies of $\mathbb{Z}_2$ and they have different roles in the group structure. When you mod out, you need to make sure you're canceling the correct group. I the case of coordinate-determined subgroups, this is easy: you cancel the corresponding coordinate. So, modding out by your $H$, and modding out by $$H':=\{(p,q,r)|p=0,q\in\mathbb{Z}_2,r\in\mathbb{Z}_{11}\}$$ give different results, despite the fact that they are isomorphic, because they sit inside the larger group differently.
