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Exercise 3.1.41 (Commutator subgroup)

Proof. Let $S = {x^{- 1} y^{- 1} 𝑥𝑦 ∣ x, y \in G}$ so that $N = ⟨ S ⟩ \leq G$ (we cannot use the results of Exercise 27, 28, 29 because we don’t know if $G$ is finite).

Let $g \in G$ . Since $γ_{g} : G \to G$ defined by $γ_{g} (x) = 𝑔𝑥 g^{- 1}$ is a (inner) automorphism,

g [x, y] g^{- 1} = γ_{g} (x^{- 1} y^{- 1} 𝑥𝑦) = γ_{g} {(x)}^{- 1} γ_{g} {(y)}^{- 1} γ_{g} (x) γ_{g} (y) = [γ_{g} (x), γ_{g} (y)]

is a commutator, so $𝑔𝑐 g^{- 1} \in N$ for every commutator $c = [x, y]$ of $G$ and every $g \in G$ .

By Proposition 9, every element $n \in N$ is of the form $n = c_{1}^{𝜀_{1}} c_{2}^{𝜀_{2}} \dots c_{n}^{𝜀_{n}}$ , where $c_{1}, c_{2}, \dots, c_{n} \in S$ are commutators, and $𝜀_{i} = \pm 1$ for each $i$ . Then for every $g \in G$ and every $n \in N$ ,

𝑔𝑛 g^{- 1} = {(g c_{1} g^{- 1})}^{𝜀_{1}} {(g c_{2} g^{- 1})}^{𝜀_{2}} \dots {(g c_{n} g^{- 1})}^{𝜀_{n}} \in N,

because the right member is a product of commutators (or inverses of commutators, which are also commutators: ${[x, y]}^{- 1} = [y^{- 1}, x^{- 1}]$ ).

Therefore the commutator subgroup is normal in $G$ .

If $\bar{x} = 𝑥𝑁$ and $\bar{y} = 𝑦𝑁$ are elements of $G ∕ N$ , where $x, y \in G$ , then $x^{- 1} y^{- 1} 𝑥𝑦 \in N$ , therefore $x^{- 1} y^{- 1} 𝑥𝑦𝑁 = N$ , thus $𝑥𝑦𝑁 = 𝑦𝑥𝑁$ or equivalently

𝑥𝑁 𝑦𝑁 = 𝑦𝑁 𝑥𝑁 .

This shows that $G ∕ N$ is abelian. □

Note: More generally, if $N \subseteq H ⊴ G$ , then $x^{- 1} y^{- 1} 𝑥𝑦 \in N \subseteq H$ , so $x^{- 1} y^{- 1} 𝑥𝑦 \in H$ . Therefore $x^{- 1} y^{- 1} 𝑥𝑦𝐻 = H$ , thus $𝑥𝑦𝐻 = 𝑦𝑥𝐻$ or equivalently

𝑥𝐻 𝑦𝐻 = 𝑦𝐻 𝑥𝐻 .

This shows that $G ∕ H$ is abelian.

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\begin{proof} 
Let $S= \{x^{-1} y^{-1} x y \mid x, y \in G\}$ so that $N = \langle S \rangle \leq G $ (we cannot use the results of Exercise 27, 28, 29 because we don't know if $G$ is finite).

Let $g \in G$. Since $\gamma_g : G \to G$ defined by $\gamma_g(x) = gxg^{-1}$ is a (inner) automorphism, 
$$g \, [x,y]\,  g^{-1} = \gamma_g(x^{-1}y^{-1} xy) = \gamma_g(x)^{-1} \gamma_g(y)^{-1} \gamma_g(x) \gamma_g(y) = [\gamma_g(x), \gamma_g(y)]$$
is a commutator, so $gcg^{-1} \in N$ for every commutator $c =  [x,y]$ of $G$ and every $g \in G$.

By Proposition 9, every element $n \in N$ is of the form $n = c_1^{\varepsilon_1}c_2^{\varepsilon_2}\cdots c_n^{\varepsilon_n}$, where $c_1, c_2,\ldots,c_n \in S$ are commutators, and $\varepsilon_i = \pm 1$ for each $i$. Then for every $g\in G$ and every $n \in N$,
$$gng^{-1} = (gc_1g^{-1})^{\varepsilon_1}(gc_2g^{-1})^{\varepsilon_2} \cdots (gc_ng^{-1})^{\varepsilon_n} \in N,$$
because the right member is a product of commutators (or inverses of commutators, which are also commutators: $[x,y]^{-1} = [y^{-1},x^{-1}]$).

Therefore the commutator subgroup is normal in $G$.

If $\overline{x} = xN$ and $\overline{y} = yN$ are elements of $G/N$, where $x,y \in G$, then $x^{-1} y^{-1} x y \in N$, therefore $x^{-1} y^{-1} x y N = N$, thus $xy N = yx N$ or equivalently 
$$xN\, yN = yN\, xN.$$
This shows that $G/N$ is abelian.
\end{proof}
Note: More generally, if $N \subseteq H \unlhd G$, then $x^{-1} y^{-1} x y \in N \subseteq H$, so $ x^{-1} y^{-1} x y \in H$. Therefore $x^{-1} y^{-1} x y H = H$, thus $xy H = yx H$ or equivalently 
$$xH\, yH = yH\, xH.$$
This shows that $G/H$ is abelian.

Exercise 3.1.41 (Commutator subgroup)

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