Structure of p-groups

Theorem

A group of order $p^2$ with prime $p$ is abelian.

\begin{proof} Assume that $Z(G)\cong {\mathbb{Z}}_p$. Then by Z theorem, $G$ is abelian. Otherwise, $Z(G)=G$ and so $G$ is abelian. \end{proof}

Theorem

Let $|G|=p^3$. Then either $G$ is abelian, or:

• for $p=2$, $G=D_8$ or $Q_8$.
• for $p⩾3$, $G={\mathbb{Z}}_{p^2}:{\mathbb{Z}}_p$ or $G=⟨a,b|a^p=b^p=c^p=1,[a,b]=c⟩$ where the exponent of $G$ equals to $p$.

\begin{proof} Assume $G$ is non-abelian. Firstly, suppose that $G$ has an element $a$ of order $p^2$. Then there exists $b\in G\backslash ⟨a⟩$ such that $G=⟨a,b⟩$. By ^qes1qo, $⟨a⟩⊲G$ and $G/⟨a⟩\simeq C_p$. If $|b|=p$, then $G=⟨a⟩:⟨b⟩$ is isomorphic to $D_8$ for $p=2$ and is isomorphic to ${\mathbb{Z}}_{p^2}:{\mathbb{Z}}_p$ for $p\ge 3$.

Otherwise, assume that $|b|=p^2$, then $b^p$ is of order $p$ and so $b^p\in ⟨a⟩$. Thus $b^p=a^{\lambda p}$ for some nonzero integer $\lambda$. When $p=2$, we have $|G|=8$ and $o(a)=o(b)=4$. Since $a^2$ is the unique element of order $2$, there is $a^2=b^2$. Since $a^b$ is of order $4$, there is $a^b=a^3$. Therefore,

$$G=⟨a,b:a^4=b^4=1,a^2=b^2,a^b=a^{-1}⟩=Q_8.$$

When $p=3$, assume that $a^b=a^j$ has order $p^2$. Notice that $Z(G)=⟨a^p⟩$, then $(a^p{)}^b=a^p$ and so $jp\equiv p(\mathrm{mod}{p}^2)$. It deduces that $j\equiv 1(\mathrm{mod}p)$ and so $a^b=a^{1+kp}$ for some $k$ with $1⩽k⩽p-1$.

We claim that there exists $\ell$ such that $o(b{a}^{-\ell })=p$. Notice that $a^p{b}^p=(ab{)}^p[b,a{]}^{p(p-1)/2}$, to compute $(b{a}^{-\ell })$, it suffices to compute $[a^{-\ell },b]$. Since

$$[a^{-\ell },b]=a^{\ell }{b}^{-1}{a}^{-\ell }{b}^{}=a^{\ell }(a^{1+kp}{)}^{-\ell }=a^{-\ell kp},$$

there is $(b{a}^{-\ell }{)}^p=b^p{a}^{-\ell p}=a^{(\lambda -\ell )p}$. Therefore, $\ell =\lambda$ is what we desired. Let $b^{\prime }=b{a}^{-\lambda }$. Since the order of $b^{\prime }$ is $p$, we get back to the last case.

Now assume that $G$ does not have elements of order $p^2$, then each elements of $G$ is of order $p$. If $p=2$, then $G$ is abelian (see here). Now assume that $p⩾3$. If $|Z(G)|=p^2$, then $G/Z(G)$ is cyclic and $G$ is abelian by Z theorem. If $|Z(G)|=p$, then $G/Z(G)$ is abelian by ^4ffac3. Then $|G^{\prime }|⩽|Z(G)|=p$ is non-trivial by $G$ non-abelian. Thus we have that $G^{\prime }=⟨c⟩={\mathbb{Z}}_p$ and $G/G^{\prime }=⟨a{G}^{\prime },b{G}^{\prime }⟩={\mathbb{Z}}_p\times {\mathbb{Z}}_p$, where $c=[a,b]$. \end{proof}

Remark.

• $G={\mathbb{Z}}_{p^2}:{\mathbb{Z}}_p$ is defined by $p_{-}^{1+2}$, and
• $G=⟨a,b|a^p=b^p=c^p=1,[a,b]=c⟩$ is defined by $p_{+}^{1+2}$. Also see extraspecial p-group.