Exercise 3.1.26 (If $S = \{x \in G \mid |x| = n\}$ and $N = \langle S \rangle$ then $N \unlhd G$.)

Proof. Let $a,b\in G$.

(a)

Since for all $g\in G$, $$(\mathit{ga}{g}^{-1})(\mathit{gb}{g}^{-1})=g(\mathit{ab})g^{-1},$$

the conjugate of the product of $a$ and $b$ is the product of the conjugate of $a$ and the conjugate of $b$ by $g$.

Moreover, for every $n\in \mathbb{Z}$,

$$a^n=1⟺{(\mathit{ga}{g}^{-1})}^n=1.$$

Since $|a|=\min <!--\; FUNCTION\; APPLICATION\; -->\{n\in {\mathbb{Z}}^{\text{+}}\mid a^n=1\}$, this shows that $|a|=|\mathit{ga}{g}^{-1}|$.

(Alternatively, the map ${\gamma }_g:G\to G$ defined by ${\gamma }_g(a)=\mathit{ga}{g}^{-1}$ is an automorphism of $G$ (inner automorphism), and every isomorphism preserves the orders of the elements.)

(b)

For all $g\in G$, $${(\mathit{ga}{g}^{-1})}^{-1}=g{a}^{-1}{g}^{-1}.$$

So the conjugate of $a^{-1}$ is the inverse of the conjugate of $a$.

(c)

Let $g$ be any element of $G$. Suppose that $\mathit{gS}{g}^{-1}\subseteq N$. Then $S\subseteq g^{-1}\mathit{Ng}$. Since $g^{-1}\mathit{Ng}$ is a subgroup which contains $S$, it contains $⟨S⟩=N$, which is by definition the smallest subgroup of $G$ containing $S$.

So $N\subseteq g^{-1}\mathit{Ng}$, or equivalently $\mathit{gN}{g}^{-1}\subseteq N$. Since this is true for every $g\in G$, this shows that

$$N⊴G.$$

(Alternatively, we may use Proposition 9: every element $n\in N$ is of the form $n=a_1^{{𝜀}_1}{a}_2^{{𝜀}_2}\cdots a_n^{{𝜀}_n}$, where $a_i\in S$. Then, since $g{a}_i{g}^{-1}\in N$ by hypothesis,

$$\mathit{gn}{g}^{-1}=g(a_1^{{𝜀}_1}{a}_2^{{𝜀}_2}\cdots a_n^{{𝜀}_n})g^{-1}={(g{a}_1{g}^{-1})}^{{𝜀}_1}{(g{a}_2{g}^{-1})}^{{𝜀}_2}\cdots {(g{a}_n{g}^{-1})}^{{𝜀}_n}\in N,$$

for all $g\in G$, so $N⊴G$.)

(d)

Suppose that $N=⟨x⟩$. By part (c) with $S=\{x\}$, $N$ is normal in $G$ if and only if for all $g\in G$, $\mathit{gx}{g}^{-1}\in N$ , that is if and only if for all $g\in G$, $\mathit{gx}{g}^{-1}=x^k$ for some integer $k$. $$⟨x⟩⊴G⟺\forall <!--\; FUNCTION\; APPLICATION\; -->g\in G,\exists <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},\mathit{gx}{g}^{-1}=x^k.$$

(e)

Let $n\in {\mathbb{Z}}^{\text{+}}$. Put $S=\{x\in G\mid |x|=n\}$ and $N=⟨S⟩$.

By part (a), for every $g\in G$ and every $x\in S$, $|\mathit{gx}{g}^{-1}|=|x|=n$, so $\mathit{gx}{g}^{-1}\in S\subseteq N$. Then by part (c),

$$N=⟨S⟩⊴G.$$