Let $G$ be a group with normal subgroup $N$, then $G/N$ is the quotient group.
If $G$ is cyclic, let its generator be $a$, then $Na$ is the generator for $G/N$. [Since $Nx = Na^i = (Na)^i$.]
If $G$ is abelian, then $G/N$ is also abelian. [Since, for $a, b \in G$, $ (a+N)(b+N)=ab+N=ba+N=(b+N)(a+N)$.]
However a more elegant method of proving these is using the first isomorphism theorem. But I am unable to come up with the mapping.
Any hints are welcome.
Assume you have some onto homomorphism $\phi$ from $G$ to $G'$ with $N$ as the kernel. We will use the fact that $G$ is abelian to show that $G'$ is also abelian. Let $a,b \in G'$. Since $\phi$ is an onto homomorphism, there exists at least one inverse image of $a$ and $b$ in $G$. Call them $x$ and $y$. So we have, $$ab = \phi(x) \phi(y) = \phi(xy) = \phi(yx) = \phi(y) \phi(x) = ba$$ Hence, for any $a, b \in G'$, $ab=ba$. Ergo, $G'$ is abelian. Now the first isomorphism theorem states that $G'$ is isomorphic with $G/N$, hence $G/N$ is also abelian.
You can use the same technique for proving the quotient qroup is cyclic.