If $M$ and $N$ are normal subgroups of $G$ and $M \subseteq N$, $G/N$ cyclic and $|N/M|=2$ prove that $G/M$ is Abelian. If $M$ and $N$ are normal subgroups of $G$ and $M \subseteq N$, $G/N$ cyclic and $|N/M|=2$ prove that $G/M$ is Abelian.
If $M$ is a subgroup of $N$, keeping in mind that $[N:M]=2$, we can get that $M \triangleleft N$. Also from Third isomorphism theorem we have: $$(G/M)/(N/M) \cong G/N$$
$G/N$ is cyclic, so $(G/M)/(N/M)$ has to be also cyclic. $G/M$ has elements of form $gM$ and $(G/M)/(N/M)$ has elements of form $gM(N/M)$ and because $N/M = \{M, nM \}$,  $gM(N/M) = \{gM, gnM \}$. This is cyclic, so $n = g^k$. If $n = g^k$ then is also $m = g^l$. I am not sure what to do with this, can someone please help me?
 A: You can also use the two following results, showing your question is part of general facts of group theory.

Lemma 1 If $N \unlhd G$ and $|N|=2$, then $N \subseteq Z(G)$

Proof If $N=\{1,n\}$ ($n \neq 1$) and $g \in G$ arbitrary, then, since $N$ is normal, $g^{-1}ng \in N$. But since $n \neq 1$, $g^{-1}ng \neq 1$, so $g^{-1}ng=n$, that is $ng=gn$, whence $n$ is central. $\square$

Lemma 2 Let $A \subseteq Z(G)$ with $G/A$ being cyclic. Then $G$ is abelian.

Proof Write $G/A=\langle \overline{g} \rangle$, where $\overline{g}$ denotes the coset $gA$. If $x,y \in G$, then $\overline{x}=\overline{g}^i$ for some integer $i$, so, $x=g^ia$, for some $a \in A$. Similarly, $y=g^jb$ for some integer $j$ and $b \in A$. Using that $A$ is central we now see, $xy=g^iag^jb=g^{i+j}ab=g^jg^iba=g^jbg^ia=yx.$ $\square$

Back to the post: we see that $N/M \unlhd G/M$ and $|N/M|=2$. By Lemma 1 we get $N/M \subseteq Z(G/M)$. But $(G/M)/(N/M) \cong G/N$ is cyclic, so by Lemma 2, $G/M$ is abelian. $\square$
A: Following Arthur's suggestion in the comments above, denote $G/M$ by $G'$ and $N/M$ as $N'$. As you've noted, $G'/N'$ is cyclic, being isomorphic to $G/N$.
Therefore, it suffices to check that every element of $N'$ commutes with every element of $G'$. Since $N'$ is normal in $G'$, for $g' \in G'$, $n_1', n_2' \in N'$, $g'n_1' = n_2'g'$. Since $N'$ is of order $2$, either $n_2' = 1$ or $n_2' = n_1'$. Suppose the former. Then $g'n_1' = g'$, hence $n_1' = 1 = n_2'$. Thus, $n_1' = n_2'$ and the elements of $N'$ commute with every element of $G'$. So $G'$ is abelian.
A: As in the other answer, we work modulo $M$ throughout to minimize the need for quotient groups, but I am also going to get rid of the prime notion (') in the other answer since that is a bit annoying to see everywhere. So rewriting $G/M$ as $G$ and $N/M$ as $N$, we are working with a group $G$ that has a normal subgroup $N$ of order $2$ such that $G/N$ is cyclic and we want to show $G$ is abelian.
Since $N$ has order $2$, we can write $N = \{1,n\}$. For $g \in G$, $gNg^{-1} = N$, which says $\{1,gng^{-1}\} = \{1,n\}$, so $gng^{-1} = n$. Thus $gn = ng$, so everything in $G$ commutes with $n$. Since $G/N$ is cyclic, let $x \in G$ represent a generator of $G/N$, so $G/N = \{x^kN : k \in \mathbf Z\}$. Thus $G = \{x^k, x^kn : k \in \mathbf Z\}$, so $G$ is generated by $x$ and $n$, which commute. Hence $G$ is abelian.
