Exercise 2.3.24 ( $(\exists a \in \mathbb{Z},\ gxg^{-1} = x^a) \iff g \in N_G(\langle x \rangle)$)

Question

Let $G$ be a finite group and let $g\in G$.
\begin{enumerate}
\item[{\bf (a)}] Prove that if $g \in N_G(\langle x \rangle)$ then $gxg^{-1} = x^a$ for some $a \in \mathbb{Z}$.
\item[{\bf (b)}] Prove conversely that if $gxg^{-1} = x^a$ for some $a\in \mathbb{Z}$ then $g \in N_G(\langle x \rangle)$.[Show first that $gx^kg^{-1} = (gxg^{-1})^k = x^{ak}$ for any integer $k$, so that $g\langle x \rangle g^{-1} \leq \langle x \rangle$. If $x$ has order $n$, show the elements $gx^ig^{-1},\ i=0,1,\ldots, n-1$ are distinct, so that $|g \langle x \rangle g^{-1} | = |\langle x \rangle | = n$ and conclude that $g \langle x \rangle g^{-1} = \langle x \rangle$.]
\end{enumerate}

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
Let $G$ be a finite group and let $g\in G$.
\begin{enumerate}
\item[{\bf (a)}] Suppose that $g \in N_G(\langle x angle)$. By definition, $g \langle x angle g^{-1} = \langle x angle$.  Since $x \in \langle x angle$, then $g x g^{-1} \in g \langle x angle g^{-1}$, so $xg^{-1} x \in \langle x angle$. Therefore there is some $a \in \Z$ such that $gxg^{-1} = x^a$.
$$g \in N_G(\langle x angle) \Rightarrow \exists a \in \Z,\ gxg^{-1} = x^a.$$

\item[{\bf (b)}]  Conversely, suppose that $gxg^{-1} = x^a$ for some $a \in \Z$. By Exercise 1.7.17, we know that 
$$
\varphi_g\ 
\left\{
\begin{array}{ccl}
G & 	o & G\
 y & \mapsto & g y g^{-1}
\end{array}
ight.
$$
is an automorphism. Therefore $\varphi(x^k) = \varphi(x)^k$ for all integers $k$, thus $gx^kg^{-1} = (gxg^{-1})^k$, so $gx^kg^{-1} = (gxg^{-1})^k = (x^a)^k =  x^{ak}$:
$$\forall k \in \Z,\ gx^k g^{-1} = x^{ak}.$$
Let $y$ be any element of $g\langle x angle g^{-1}$. Then $y = gx^k g^{-1}$ for some integer $k$, thus $y = x^{ak}$, so $y \in \langle x angle$. This shows that
$$g \langle x angle g^{-1} \leq \langle x angle.$$

Suppose now that $|x| =n$.
Since for all $y \in \langle x angle$, $gyg^{-1} \in \langle x angle$, we may consider the restriction $\psi_g$ of $\varphi_g$, defined by
$$
\psi_g\ 
\left\{
\begin{array}{ccl}
\langle x angle & 	o & \langle x angle\
 y & \mapsto & \psi_g(y) = \varphi_g(y) = g y g^{-1}
\end{array}
ight.
$$
 Since $\varphi_g$  is an automorphism,
\begin{itemize} 
\item $\psi_g$ is a homomorphism.

\item $\psi_g$ is injective : for all $y,z\in \langle x angle$, $\psi_g(y)= \psi_g(z) \Rightarrow \varphi_g(x) = \varphi_g(y) \Rightarrow x = y$.

(Less formally, this means that if $ \langle x angle = \{1,x,x^2,\ldots,x^{n-1}\}$, then  the elements $gx^i g^{-1},\ i= 0,1,\ldots,n-1$, are distinct.)
\item Since $\psi_g : \langle x angle  	o  \langle x angle$ is injective, and $\langle x angle$ is a finite set, $\psi_g$ is surjective.
\end{itemize}
Therefore $\psi_g$ is an automorphism of $\langle x angle$. In particular $\psi_g( \langle x angle) =  \langle x angle$, so
 $$g  \langle x angle g^{-1} =  \langle x angle.$$
 This shows that $g \in N_G(\langle x angle)$.
 
 \bigskip
 In conclusion, for all $g,x \in G$,
  $$(\exists a \in \mathbb{Z},\ gxg^{-1} = x^a) \iff g \in N_G(\langle x angle).$$

\end{enumerate}
\end{proof}

Exercise 2.3.24 ( $(\exists a \in \mathbb{Z},\ gxg^{-1} = x^a) \iff g \in N_G(\langle x \rangle)$)

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Exercise 2.3.24 ( $(\exists a \in \mathbb{Z},\ gxg^{-1} = x^a) \iff g \in N_G(\langle x \rangle)$)

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