Write $G/Z(G)=\langle \bar{g} \rangle$ for some $g \in G$.

Any element $x\in G$ can be written as $x=g^a z$ for some $z \in Z(G)$ and $a \in \Z$.

Using this expression, show that $xy=yx$ for any $x, y \in G$.

Proof.

Since the quotient group $G/Z(G)$ is cyclic, it is generated by one element.
Let $g \in G$ be an element such that $\bar{g}=gZ(G)$ is a generator of $G/Z(G)$. Namely, $\langle \bar g \rangle =G/ Z(G)$.

Then for any element $x \in G$, we have $\bar{x}\in G/Z(G)=\langle \bar g \rangle$ and hence $\bar{x}=\bar{g}^a$ for some $a \in \Z$.
It follows that any element of $G$ can be written as $x=g^a z$ for some $z \in Z(G)$ and $a \in \Z$.

Take any two elements $x, y \in G$ and write $x=g^a z$ and $y=g^b w$, where $a, b \in \Z$ and $z, w \in Z(G)$.

Then we claim that $xy=yx$. To see this we calculate as follows.
\begin{align*}
xy&= g^a z g^b w\\
&=g^a g^b z w && \text{since $z\in Z(G)$}\\
&= g^{a+b} wz && \text{since $w\in Z(G)$}\\
&=g^bg^awz\\
&=g^b w g^az&& \text{since $w\in Z(G)$}\\
&=yx.
\end{align*}

Since the product of any two elements of $G$ is commutative, we conclude that $G$ is abelian.

Group of Order $pq$ is Either Abelian or the Center is Trivial
Let $G$ be a group of order $|G|=pq$, where $p$ and $q$ are (not necessarily distinct) prime numbers.
Then show that $G$ is either abelian group or the center $Z(G)=1$.
Hint.
Use the result of the problem "If the Quotient by the Center is Cyclic, then the Group is […]

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Suppose the order of a group $G$ is $p^2$, where $p$ is a prime number.
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Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of […]

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Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]

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Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism.
Prove that we have an isomorphism of groups:
\[G \cong \ker(f)\times \Z.\]
Proof.
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Definition.
A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]

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Prove that every cyclic group is abelian.
Proof.
Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.)
Let $a$ and $b$ be arbitrary elements in $G$.
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Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
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[…] of the quotient $|G/Z(G)|=[G:Z(G)]=p$ is a prime, thus $G/Z(G)$ is a cyclic group. Recall that if the quotient by the center is cyclic, then the group is abelian. Thus the group $G$ is abelian, which again a […]

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[…] Use the result of the problem If the quotient by the center is cyclic, then the group is abelian. […]

[…] of the quotient $|G/Z(G)|=[G:Z(G)]=p$ is a prime, thus $G/Z(G)$ is a cyclic group. Recall that if the quotient by the center is cyclic, then the group is abelian. Thus the group $G$ is abelian, which again a […]

[…] If the quotient by the center is cyclic, then the group is abelian (post 2) […]