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.
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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 […]

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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 $\sigma(i)=j$
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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 […]

Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$
Let $D_8$ be the dihedral group of order $8$.
Using the generators and relations, we have
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]
(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$.
Prove that the centralizer […]

[…] 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 […]