Since $G$ is a $p$-group, its center is not trivial (see post 1 for a proof.)

If the center $Z(G)=G$, then $G$ is abelian so assume that $Z(G)$ is a proper nontrivial subgroup. Then the center must have order $p$ and it follows that the order of the quotient $G/Z(G)$ is $p$, hence $G/Z(G)$ is a cyclic group.

(b) The group $G$ is isomorphic to either $\Zmod{p^2}$ or $\Zmod{p} \times \Zmod{p}$

Let $x \in G$ be any nontrivial element of $G$. If $x \in G$ has order $p^2$, then $G=\langle x \rangle \cong \Zmod{p^2}$. If the order of $x$ is $p$, then take $y \in G \setminus \langle x \rangle$. Then the order of $y$ is also $p$.

Since the subgroup $\langle x, y \rangle$ generated by $x$ and $y$ is properly bigger than the subgroup $\langle x \rangle$, we must have $G=\langle x, y \rangle$.
We claim that $\langle x, y \rangle \cong \langle x \rangle \times \langle y \rangle$.

Define a map $f:\langle x \rangle \times \langle y \rangle \to \langle x, y \rangle$ by sending $(x^a, y^b)$ to $x^ay^b$. This is a group homomorphism because for any elements $(x^{a_1}, y^{b_1})$ and $(x^{a_2}, y^{b_2})$, we have
\begin{align*}
f\left( (x^{a_1}, y^{b_1})(x^{a_2}, y^{b_2}) \right) &=f\left (x^{a_1+a_2}, y^{b_1+b_2})\right) =x^{a_1+a_2} y^{b_1+b_2}\\
& =x^{a_1}y^{b_1}x^{a_2}y^{b_2} =f\left( (x^{a_1}, y^{b_1}) \right) f\left( (x^{a_2}, y^{b_2}) \right).
\end{align*}

Here we used the result of part (a) that $G$ is abelian in the third equality.

We claim that the homomorphism $f$ is injective.
If $f\left (x^a, y^b) \right)=1$, we have $x^a=y^b$ but since $y \not \in \langle x \rangle$ we must have $a=b=0$. Thus the kernel is trivial, hence $f$ is injective.

Since $\langle x \rangle \times \langle y \rangle \cong \Zmod{p} \times \Zmod{p}$ has order $p^2$ and $f$ is injective, the homomorphism must be surjective as well, hence it is an isomorphism.

Abelian Normal subgroup, Quotient Group, and Automorphism Group
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 […]

The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
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 […]

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 […]

Group Homomorphism, Conjugate, Center, and Abelian group
Let $G$ be a group. We fix an element $x$ of $G$ and define a map
\[ \Psi_x: G\to G\]
by mapping $g\in G$ to $xgx^{-1} \in G$.
Then prove the followings.
(a) The map $\Psi_x$ is a group homomorphism.
(b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the […]

If the Quotient by the Center is Cyclic, then the Group is Abelian
Let $Z(G)$ be the center of a group $G$.
Show that if $G/Z(G)$ is a cyclic group, then $G$ is abelian.
Steps.
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 […]

Fundamental Theorem of Finitely Generated Abelian Groups and its application
In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Problem.
Let $G$ be a finite abelian group of order $n$.
If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]

The Center of the Symmetric group is Trivial if $n>2$
Show that the center $Z(S_n)$ of the symmetric group with $n \geq 3$ is trivial.
Steps/Hint
Assume $Z(S_n)$ has a non-identity element $\sigma$.
Then there exist numbers $i$ and $j$, $i\neq j$, such that $\sigma(i)=j$
Since $n\geq 3$ there exists another […]

Equivalent Definitions of Characteristic Subgroups. Center is Characteristic.
Let $H$ be a subgroup of a group $G$. We call $H$ characteristic in $G$ if for any automorphism $\sigma\in \Aut(G)$ of $G$, we have $\sigma(H)=H$.
(a) Prove that if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$.
(b) Prove that the center […]

[…] order of $P$ is $9$, a square of a prime number, thus $P$ is abelian. (See A group of order the square of a prime is abelian.) Also, the order of the quotient group $G/P$ is $2$, thus $G/P$ is an abelian (cyclic) group. Thus […]

[…] these Sylow subgroups are of order $11^2$ and $13^2$, respectively, they are abelian. Since the direct product of abelian groups is abelian, the group $G$ is […]

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[…] order of $P$ is $9$, a square of a prime number, thus $P$ is abelian. (See A group of order the square of a prime is abelian.) Also, the order of the quotient group $G/P$ is $2$, thus $G/P$ is an abelian (cyclic) group. Thus […]

[…] these Sylow subgroups are of order $11^2$ and $13^2$, respectively, they are abelian. Since the direct product of abelian groups is abelian, the group $G$ is […]