Problem 5.4.9 (Kernel of $x \mapsto x^p$ in a $p^3$-group)

Question

Prove that if $p$ is an odd prime and $P$ is a group of order $p^3$ then the $p^{\mathrm{th}}$ power map $x \mapsto x^p$ is a homomorphism of $P$ into $Z(P)$. If $P$ is not cyclic, show that the kernel of the $p^{\mathrm{th}}$ power map has order $p^2$ or $p^3$. Is the squaring map a homomorphism in non-abelian groups of order $8$? Where the oddness of $p$ needed in the above proof? [Use Exercise 8.]

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $\varphi$ be defined by
$$\varphi
\left\{
\begin{array}{ccl}
P & 	o & P\
x & \mapsto &x^p
\end{array}
ight.
$$
We first prove that $\varphi$ is a homomorphism. This is obviously true if $P$ is abelian. 

Now we suppose that $P$ is not abelian.  
By Exercise 7, $P' = Z(P)$, thus $[x,y] \in Z(P)$ for all $x,y \in P$. Moreover $|Z(P)| = p$ (see the proof of Exercise 7). Therefore 
\begin{align*}
[x,y]^p = 1\qquad (x \in P, \ y\in P).
\end{align*}
Since $p$ is odd, $[x,y]^{p\left(\frac{p-1}{2}ight)} = 1$. Since $[x,y] \in Z(P)$, then $x$ and $y$ commute with $[x,y]$, thus, by Exercise 8,
\begin{align*}
(xy)^p &= x^p y^p [y,x]^{\frac{p(p-1)}{2}}\
&= x^p y^p \left([x,y]^{\frac{p(p-1)}{2}}ight)^{-1}\
&= x^p y^p.
\end{align*}
This shows that $\varphi$ is a homomorphism.

Now, by  (1), and  by equality (2) in the proof of Exercise 8, for all $x,y \in P$,
$$[x^p ,y] = [x,y]^p = 1,$$
 thus $x^p y = y x^p$, so $x^p \in Z(P)$. This shows that $\mathrm{im}(\varphi) \subseteq Z(P)$, so we may consider the map
$$\psi
\left\{
\begin{array}{ccl}
P & 	o & Z(P)\
x & \mapsto &x^p
\end{array}
ight.
$$
which is well defined, and is a homomorphism, since $x^p y^p = (xy)^p$ for all $x,y \in P$, as proven above.

If $P$ is abelian, $Z(P) = P$ so $\psi$ is a well defined homomorphism.

In both cases ($P$ abelian or not abelian), $\psi$ is a well defined homomorphism.

\begin{itemize}
\item If $P$ is not abelian, then $|Z(P)| = p$. Moreover
$$|P/\ker(\varphi)| = |\mathrm{im}(\varphi)| 	ext{ divides } p = |Z(P)|,$$
therefore $p^2$ divides $ |\ker(\varphi)|$, where $|\ker(\varphi)|$ divides $p^3$, so
$$|\ker(\varphi)| = p^2 	ext{ or } p^3.$$

\item If $P$ is abelian (but not cyclic by hypothesis), then
$$P \simeq Z_p^2 	imes Z_p \qquad 	ext{or} \qquad P \simeq Z_p 	imes Z_p 	imes Z_p.$$
    \begin{itemize}
    \item If $P \simeq Z_p^2 	imes Z_p$ the rank of $P$ is $n = 2$. By  Exercise 5.2.7,
    $$\ker(\varphi) \simeq E_{p^2},$$
    the elementary $p$-group of order $p^2$.
    \item If $P \simeq Z_p	imes Z_p 	imes Z_p$ the rank of $P$ is $n = 3$. By  Exercise 5.2.7,
    $$\ker(\varphi) \simeq E_{p^3}.$$

    \end{itemize}
    In both cases (rank 2 or rank 3), 
    $$|\ker(\varphi)| = p^2 	ext{ or } p^3.$$
\end{itemize}
In conclusion, if $P$ is any non cyclic group of order $p^3$ (abelian or not), then 
    $$|\ker(\varphi)| = p^2 	ext{ or } p^3.$$
    
\bigskip
In the quaternionic group $Q_8$,
$$(ij)^2 =k^2 = -1 
e i^2 j^2 =  1,$$
so the the squaring map is not a homomorphism in the non-abelian groups $Q_8$ of order $8$ (similarly, in $D_8$, $(rs)^2 = 1 
e r^2s^2 = r^2$).

We use the oddness of $p$ in the above proof when we write
\begin{align*}
 [x,y] ^{\frac{p(p-1)}{2}} = \left([x,y]^{p} ight)^{\frac{p-1}{2}} = 1.
 \end{align*}
\end{proof}

Problem 5.4.9 (Kernel of $x \mapsto x^p$ in a $p^3$-group)

Answers

Comments

Problem 5.4.9 (Kernel of $x \mapsto x^p$ in a $p^3$-group)

Answers

Comments

Add answer