Are these two quotient groups of $U(16)$ isomorphic? Let $G=U(16)$, $H=\{1,15\}$, $K=\{1,9\}$.
Are $H$ and $K$ isomorphic? What about $G/H$ & $G/K$?
I found that $H$ and $G$ are isomorphic as both are cyclic and every finite cyclic group is isomorphic to group $Z_n$. 
Here both $H$ and $G$ are isomorphic to  $Z_2$. So both $H$ and $G$ are isomorphic. Now as $H$ and $G$ are cyclic, can I conclude that the quotient groups are abelian?
 A: (Note that almost every time you mention $H$ and $G$, you really mean $H$ and $K$. I will continue to refer to the two subgroups as $H$ and $K$).
As $U(16)$ is abelian, any subgroup is abelian. This includes quotient groups.
The group $U(16) = \{1,3,5,7,9,11,13,15\}$ and has eight elements. As $H$ and $K$ each have $2$ elements, we know that $G/H$ and $G/K$ will each have $4$ elements. Let's find the quotient groups explicitly, as it's a good exercise in understanding quotient groups.
First, let's find $G/H$.
$$ \begin{array}{lll}
1\cdot \{1,15\} &= \{1,15\} &= 15 \cdot \{1,15\} \\
3\cdot \{1,15\} &= \{3,13\} &= 13 \cdot \{1,15\} \\
5 \cdot\{1,15\} &= \{5,11\} &= 11 \cdot \{1,15\} \\
7 \cdot \{1,15\} &= \{7,9\} &= 9 \cdot \{1,15\} 
\end{array}$$ 
This group is cyclic of order $4$, and the coset $3 \cdot \{1,15\}$ is a generator.
Now let's find $G/K$.
$$\begin{array}{lll}
1\cdot\{1,9\} &= \{1,9\} &= 9\cdot\{1,9\} \\
3\cdot\{1,9\} &= \{3,11\} &= 11\cdot\{1,9\} \\ 
5\cdot\{1,9\} &= \{5,13\} &= 13\cdot\{1,9\} \\
7 \cdot \{1,9\} &= \{7, 15\} &= 15\cdot\{1,9\}
\end{array}$$
Notice that $(3 \cdot\{1, 9\})^2 = 9 \cdot \{1, 9\} = \{1,9\}$, and $(5 \cdot\{1,9\})^2 = 9 \cdot \{1,9\} = \{1,9\}$, and $(7\cdot\{1,9\})^2 = 1 \cdot \{1,9\} = \{1,9\}$. So the square of every coset is the identity.
Thus $G/K$ is not cyclic, and so the two quotient groups are not isomorphic. (In fact, note that in $G/H$, we have that $3 \{1,9\}$ times $5 \{1,9\}$ is $15\{1,9\} = 7\{1,9\}$, so in fact $G/H$ is isomorphic to the Klein four group, not the cyclic group). $\spadesuit$
