This is a homework problem for an abstact Algebra course.
Identify the group $(28\mathbb Z + 20\mathbb Z)/20 \mathbb Z$.
I did this just by looking at individual elements of $(28\mathbb Z + 20\mathbb Z)$, and what happens to them after modding out by $20$.
Elements of the "numerator" (not sure the right word) take the form $28i + 20j$ for integers $i,j$. However, since we are modding out by $20$, the additive factor of $20j$ doesn't change things. Therefore, by looking at $28i$ for $i\in \mathbb Z$, we see that
$28 \equiv 8 \mod 20$,
$56 \equiv 16 \mod 20$,
$84 \equiv 4 \mod 20$, etc.
So we get that this group looks like $\{0,4,8,12,16\}$.
Equivalently, I think this looks like $4\mathbb Z / 20 \mathbb Z$. In other words, multiples of $4$, except multiples of $20$ are "equal to zero."
In general, I possibly see a pattern that $n \mathbb Z / m \mathbb Z$ is given by $\gcd(m,n)\mathbb Z / m \mathbb Z$.
Is this correct?
(since this is homework, hints or corrections are also fine, instead of a full answer).
When $H,K\le G$ and one of them is normal, $HK$ is a subgroup.
The only subgroups of $\Bbb Z$ are the cyclic $n\Bbb Z.$
It's easy to see that $20\Bbb Z+28\Bbb Z\cong 4\Bbb Z$. For we have that there's $a,b$ such that $a20+b28=4,$ by Bezout.
So we have $4\Bbb Z/20\Bbb Z\cong \Bbb Z/5\Bbb Z.$