[proofplan] We prove normality by showing that conjugation by an arbitrary element of $G$ preserves the commutator subgroup. The key computation is that the conjugate of a commutator is again a commutator, with both entries conjugated. Since $[G,G]$ is generated by all commutators, this preservation of generators implies preservation of every finite product of generators and their inverses. [/proofplan] [step:Identify the generators of the commutator subgroup] For $a,b \in G$, use the commutator convention \begin{align*} [a,b] = aba^{-1}b^{-1}. \end{align*} Define the set of commutators $S \subset G$ by \begin{align*} S = \{[a,b] : a,b \in G\}. \end{align*} By definition of the commutator subgroup, \begin{align*} [G,G] = \langle S \rangle. \end{align*} Thus an element of $[G,G]$ is a finite product of elements of $S$ and inverses of elements of $S$. [/step] [step:Show that conjugation sends each commutator to a commutator] Fix $x \in G$. Define the conjugation map \begin{align*} c_x: G &\to G \end{align*} \begin{align*} y &\mapsto xyx^{-1}. \end{align*} Let $a,b \in G$. Then \begin{align*} c_x([a,b]) = xaba^{-1}b^{-1}x^{-1}. \end{align*} Using associativity and the identities $x^{-1}x=e$, $xa^{-1}x^{-1}=(xax^{-1})^{-1}$, and $xb^{-1}x^{-1}=(xbx^{-1})^{-1}$, we compute \begin{align*} xaba^{-1}b^{-1}x^{-1} = (xax^{-1})(xbx^{-1})(xa^{-1}x^{-1})(xb^{-1}x^{-1}). \end{align*} Therefore \begin{align*} c_x([a,b]) = [xax^{-1}, xbx^{-1}]. \end{align*} Since $xax^{-1}, xbx^{-1} \in G$, the element $[xax^{-1}, xbx^{-1}]$ lies in $S$. Hence $c_x(S) \subset S$. [guided] Fix an arbitrary element $x \in G$. To prove normality, we must understand what conjugation by $x$ does to the elements that generate $[G,G]$. Define the conjugation map \begin{align*} c_x: G &\to G \end{align*} \begin{align*} y &\mapsto xyx^{-1}. \end{align*} Now take arbitrary $a,b \in G$. Their commutator is \begin{align*} [a,b] = aba^{-1}b^{-1}. \end{align*} Conjugating this element gives \begin{align*} c_x([a,b]) = xaba^{-1}b^{-1}x^{-1}. \end{align*} The goal is to rewrite this expression as a commutator. We insert the cancelling factors $x^{-1}x=e$ between the four letters $a$, $b$, $a^{-1}$, and $b^{-1}$: \begin{align*} xaba^{-1}b^{-1}x^{-1} = (xax^{-1})(xbx^{-1})(xa^{-1}x^{-1})(xb^{-1}x^{-1}). \end{align*} The inverse of $xax^{-1}$ is $xa^{-1}x^{-1}$, because \begin{align*} (xax^{-1})(xa^{-1}x^{-1}) = xaa^{-1}x^{-1} = e, \end{align*} and similarly the inverse of $xbx^{-1}$ is $xb^{-1}x^{-1}$. Therefore the last display is exactly the commutator of $xax^{-1}$ and $xbx^{-1}$: \begin{align*} c_x([a,b]) = [xax^{-1}, xbx^{-1}]. \end{align*} Because $xax^{-1}$ and $xbx^{-1}$ are elements of $G$, this is again an element of the commutator set $S=\{[u,v]:u,v\in G\}$. Thus conjugation by $x$ sends every commutator generator to another commutator generator. [/guided] [/step] [step:Extend conjugation invariance from generators to the subgroup they generate] Let $h \in [G,G]$. If $h=e$, then \begin{align*} c_x(h)=xex^{-1}=e \in [G,G]. \end{align*} Assume now that $h \neq e$. Since $[G,G]=\langle S\rangle$, there exist a positive integer $m$, elements $s_1,\dots,s_m \in S$, and signs $\varepsilon_1,\dots,\varepsilon_m \in \{1,-1\}$ such that \begin{align*} h=s_1^{\varepsilon_1}\cdots s_m^{\varepsilon_m}. \end{align*} The map $c_x$ is a [group homomorphism](/page/Group%20Homomorphism), since for all $u,v \in G$, \begin{align*} c_x(uv)=xuvx^{-1}=(xux^{-1})(xvx^{-1})=c_x(u)c_x(v). \end{align*} Therefore \begin{align*} c_x(h)=c_x(s_1)^{\varepsilon_1}\cdots c_x(s_m)^{\varepsilon_m}. \end{align*} By the previous step, each $c_x(s_i)$ lies in $S \subset [G,G]$. Since $[G,G]$ is a subgroup, it contains inverses and finite products of its elements. Hence $c_x(h)\in [G,G]$. [/step] [step:Conclude that the commutator subgroup is normal] We have shown that for every $x \in G$ and every $h \in [G,G]$, \begin{align*} xhx^{-1}=c_x(h)\in [G,G]. \end{align*} Thus \begin{align*} x[G,G]x^{-1}\subset [G,G] \end{align*} for every $x \in G$. Since $[G,G]$ is a subgroup of $G$, this is precisely the conjugation-invariance criterion for normality. Therefore \begin{align*} [G,G]\trianglelefteq G. \end{align*} [/step]