Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.24
Let $G$ be a finite group and let $x \in G$.
(1) Prove that if $g \in N_G(\langle x \rangle)$ then $gxg^{-1} = x^a$ for some integer $a$.
(2) Show conversely that if $gxg^{-1} = x^a$ for some integer $a$, then $g \in N_G(\langle x \rangle)$. [Hint: Show first that $$gx^kg^{-1} = (gkg^{-1})^k = x^{ak}$$ for any integer $k$, so that $g \langle x \rangle g^{-1} \leq \langle x \rangle$. If $x$ has order $n$, show that the elements $gx^ig^{-1}$ are distinct for $i \in \{ 0, 1, \ldots, n-1 \}$, so that $|g \langle x \rangle g^{-1}| = |\langle x \rangle| = n$ and conclude that $g \langle x \rangle g^{-1} = \langle x \rangle$.]
Solution:
(1) Let $g \in N_G(\langle x \rangle)$. By definition, we have $gxg^{-1} \in \langle x \rangle$, so that $gxg^{-1} = x^a$ for some integer $a$.
(2) We prove some lemmas.
Lemma 1: Let $G$ be a group and let $x,g \in G$. Then for all integers $k$, $gx^kg^{-1} = (gxg^{-1})^k$.
Proof: First we prove the conclusion for nonnegative $k$ by induction on $k$. If $k = 0$, we have $$gx^0g^{-1} = gg^{-1} = 1 = (gxg^{-1})^0.$$ Now suppose the conclusion holds for some $k \geq 0$; then $$gx^{k+1}g^{-1} = gxx^kg^{-1} = gxg^{-1}gx^kg^{-1} = gxg^{-1}(gxg^{-1})^k = (gxg^{-1})^{k+1}.$$ By induction, the conclusion holds for all nonnegative $k$. Now suppose $k < 0$; then $$gx^kg^{-1} = (gx^{-k}g^{-1})^{-1} = (gxg^{-1})^{-k^{-1}} = (gxg^{-1})^k.$$ Thus the conclusion holds for all integers $k$. $\square$
Lemma 2: Let $G$ be a group and let $x,g \in G$ such that $gxg^{-1} = x^a$ for some integer $a$. Then $g \langle x \rangle g^{-1}$ is a subgroup of $\langle x \rangle$.
Proof: Let $gx^kg^{-1} \in g \langle x \rangle g^{-1}$; by Lemma 1 we have $gx^kg^{-1} = (gxg^{-1})^k = x^{ak}$, so that $gxg^{-1} \in \langle x \rangle$. Thus $g \langle x \rangle g^{-1} \subseteq \langle x \rangle$. Now let $gx^bg^{-1}$, $gx^cg^{-1} \in g \langle x \rangle g^{-1}$. Then $$gx^bg^{-1}(gx^cg^{-1})^{-1} = gx^bg^{-1}gx^{-c}g^{-1} = gx^{b-c}g^{-1} \in g \langle x \rangle g^{-1}.$$ By the Subgroup Criterion, then, $g \langle x \rangle g^{-1} \leq \langle x \rangle$. $\square$
Lemma 3: Let $G$ be a group and let $x,g \in G$ such that $gxg^{-1} = x^a$ for some integer $a$ and such that $|x| = n$, $n \in \mathbb{Z}$. Then $gx^ig^{-1}$ are distinct for $i \in \{ 0, 1, \ldots, n-1 \}$.
Proof: Choose distinct $i,j \in \{ 0, 1, \ldots, n-1 \}$. By a previous exercise, $x^i \neq x^j$. Suppose now that $gx^ig^{-1} = gx^jg^{-1}$; by cancellation we have $x^i = x^j$, a contradiction. Thus the $gx^ig^{-1}$ are distinct. $\square$
Now to the main result; suppose $gxg^{-1} = x^a$ for some integer $a$. Since $G$ has finite order, $|x| = n$ for some $n$. By Lemma 2, $g \langle x \rangle g^{-1} \leq \langle x \rangle$, and by Lemma 3 we have $|g \langle x \rangle g^{-1}| = |\langle x \rangle|$. Since $G$ is finite, then, we have $g \langle x \rangle g^{-1} = \langle x \rangle$. Thus $g \in N_G(\langle x \rangle)$.
