Can you give example of application of theorem "if $G/Z(G)$ is cyclic, then $G$ is Abelian?" For $G$ a group and $Z(G)$ the center of $G$, then proving "if $G/Z(G)$ is cyclic, then $G$ is Abelian" is a standard undergraduate problem. I have not found any examples of this theorem. Can you give me some?
 A: The following would be an interesting example, as it says some interesting property of an example of a group.
There is no group $G$ such that $G/Z(G)$ is isomorphic to quaternion group $Q_8$.
Suppose $G$ is a group such that $G/Z(G)$ is quaternion group $Q_8=\langle i,j\rangle$. 
Note that $G/Z(G)=Q_8$ can be written as product of two cyclic subgroups $\langle i\rangle$, $\langle j\rangle$ (order $4$). 
Correspondingly, $G$ can be written as product of two subgroups $H_1$, $H_2$ (which contain $Z(G)$).
Then $H_1/Z(G)=\langle i\rangle$, i.e. $H_1/Z(G)$ is cyclic, hence $H_1$ is abelian. Similarly $H_2$ is abelian.
Thus, $G$ is product of two abelian subgroups $H_1$ and $H_2$. 
Next, in $G/Z(G)$, which is $Q_8$, note that $\langle i\rangle \cap \langle j\rangle $ is non-trivial (it is $\{1,-1\}$). Hence, correspondingly, in $G$, $H_1\cap H_2$ is not $Z(G)$; it is bigger than $Z(G)$.
Consider $x\in H_1\cap H_2$ such that $x\notin Z(G)$.
Since $x$ commutes with all elements of $H_1$ as well as $H_2$ (why?), and since $G$ is product of $H_1$ and $H_2$, it follows that $x$ commutes with all the elements of $G$, i.e. $x\in Z(G)$. Contradiction!!!
Thus there is no group $G$ such that $G/Z(G)$ is quaternion group of order $8$.
A: You can't have $O(\frac {G}{Z(G)})=15,35... $ to be precise no number of the form $pq$ where p,q are distinct prime such that $p$ doesn't divide $q-1$.Since groups with this kind of order are always cyclic.
A: Theorem. There is no group $G$ such that $\operatorname{Aut}(G)\cong\mathbb{Z}$.
Proof. Suppose $G$ is such that $\operatorname{Aut}(G)\cong\mathbb{Z}$. Then $G/Z(G)\cong\operatorname{Inn}(G)$ and $\operatorname{Inn}(G)\leq\operatorname{Aut}(G)$ so $G/Z(G)$ is cyclic. Hence $G$ is abelian.
Every abelian group $G$ has an automorphism of the form $\phi: g\mapsto g^{-1}$ for all $g\in G$, and this must be trivial (as otherwise it has order $2$ but $\operatorname{Aut}(G)$ has no elements of order $2$). Hence, every element of $G$ has order $2$. Then $G$ must be the direct sum of cyclic groups of order two. The factors in this sum can be permutated giving non-trivial automorphisms of finite order, a contradiction. Hence, $\operatorname{Aut}(G)\not\cong \mathbb{Z}$ as required.

This argument actually uses choice (to form the direct sum). If you do not wish to assume choice then you need to restate the theorem: There is no finitely generated $G$ such that... I have a faint memory that someone called Asaf Karagila proved that if you do not assume choice then there exists some universe containing a group $G$ with $\operatorname{Aut}(G)\cong\mathbb{Z}$.
A: Here is another application/generalization - let $1=\zeta_0(G) \subseteq Z(G)=\zeta_1(G) \subseteq \zeta_2(G) \subseteq \cdots$ denote the upper central series of $G$ and $G=\gamma_1(G) \supseteq \gamma_2(G)=[G,G] \supseteq \gamma_3(G)=[\gamma_2(G),G] \supseteq \cdots$ the lower central series of $G$.
Proposition Let $n \geq 1$. If $G/\zeta_n(G)$ is cyclic then $\gamma_{n+1}(G)=1$ (equivalently: $G$ is nilpotent of class $n$).
Proof $G/\zeta_n(G)=(G/\zeta_{n-1}(G))/Z(G/\zeta_{n-1}(G))$, whence $G/\zeta_{n-1}(G)$ is abelian, that is, $\gamma_2(G) \subseteq \zeta_{n-1}(G)$. But then $\gamma_3(G)=[\gamma_2(G),G] \subseteq [\zeta_{n-1}(G),G] \subseteq \zeta_{n-2}(G)$. Proceeding like this, the result follows.
