Definition 2.2.1 Let $U(G)$ be the set of commutators of the group $G$: $$U(G)=\{xyx^{-1}y^{-1}|x,y\in G\}.$$ The commutator subgroup of $G$, denoted $G'$, is the smallest subgroup containing $U(G)$, that is, $G'=\langle U(G)\rangle$. $\square$
Lemma 2.2.2 If $X$ is any subset of $G$ and for all $g\in G$, $gXg^{-1}\subseteq X$, then $\langle X\rangle$ is a normal subgroup of $G$.
Proof. Note that the hypothesis implies that for all $g\in G$, $X\subseteq g^{-1}Xg$. Suppose that $y\in \langle X\rangle$; we need to show that for all $g\in G$, $gyg^{-1}\in \langle X\rangle$. Let $H=g^{-1}\langle X\rangle g$; $H$ is a subgroup of $G$, called a conjugate of $\langle X\rangle$ (this is an easy exercise). Since $H=g^{-1}\langle X\rangle g \supseteq g^{-1}Xg \supseteq X$, $H$ is a subgroup containing $X$ and therefore $H\supseteq \langle X\rangle$, so $y\in H$. Now $gyg^{-1}\in gHg^{-1}=gg^{-1}\langle X\rangle gg^{-1} =\langle X\rangle$. $\qed$
Proof. It suffices to check that $gU(G)g^{-1}\subseteq U(G)$, that is, that for any $x$, $y$, and $g$ in $G$, $gxyx^{-1}y^{-1}g^{-1}$ is a commutator. Write $$\eqalign{ gxyx^{-1}y^{-1}g^{-1}&=(gxg^{-1})(gyg^{-1})(gx^{-1}g^{-1})(gy^{-1}g^{-1})\cr &=uvu^{-1}v^{-1},\cr }$$ where $u=gxg^{-1}$ and $v=gyg^{-1}$. $\qed$
Lemma 2.2.4 $G/G'$ is abelian.
Proof. We need to show that $xyG'=yxG'$ for all $x$ and $y$ in $G$. By properties of cosets, this is true if and only if $(yx)^{-1}xy\in G'$. This is true because $(yx)^{-1}xy=x^{-1}y^{-1}xy\in U(G)\subseteq G'$. $\qed$
Proof. For any $x$ and $y$ in $G$, $xyH=yxH$, so $x^{-1}y^{-1}xyH=H$ or $x^{-1}y^{-1}xy\in H$—that is, $H$ contains every commutator, so $H\supseteq G'$. $\qed$
Definition 2.2.6 $G^{(0)}=G$ and $G^{(k)}=\left(G^{(k-1)}\right)'$ for $k\ge 1$. $\square$
Theorem 2.2.7 $G$ is solvable iff $G^{(k)}=\{e\}$ for some $k$.
Proof. If $G^{(k)}=\{e\}$ then $\{e\}=G^{(k)}\nsg G^{(k-1)} \nsg\cdots\nsg G''\nsg G'\nsg G$ and $G^{(i-1)}/G^{(i)}$ is abelian.
Suppose $\{e\}=N_k\nsg N_{k-1}\nsg\cdots\nsg N_1\nsg N_0=G$ and $N_i/N_{i+1}$ is abelian. By lemma 2.2.5, $N_1\supseteq G'$, $N_2\supseteq N_1'\supseteq G''$, and by an easy induction, $N_k\supseteq G^{(k)}$, so $\{e\}=G^{(k)}$. $\qed$
Lemma 2.2.8 $G^{(i)}\nsg G$ for all $i$.
Proof. Suppose that $N\nsg G$; we prove that $N'\nsg G$. Suppose $g\in G$ and $u\in N'$; we need to prove that $gug^{-1}\in N'$. By lemma 2.2.2, it suffices to prove that if $u\in U(N)$ then $gug^{-1}\in U(N)$. This is almost identical to the proof of lemma 2.2.3. Now an easy induction completes the proof. $\qed$
Lemma 2.2.9 Let $G=S_n$, $n\ge 5$. For every $k$, $G^{(k)}$ contains every 3-cycle in $S_n$.
Proof. Suppose $N\nsg G$ and $N$ contains all 3-cycles, so $N'$ contains the commutator $(123)(145)(321)(541)=(124)$. $N'$ is normal in $G$, so for all $\pi\in S_n$, $\pi(124)\pi^{-1} \in N'$. Choose $\pi$ so that $\pi(1)=i$, $\pi(2)=j$, $\pi(4)=k$. Then $(ijk)=\pi(124)\pi^{-1}$, so every 3-cycle is in $N'$.
Now we prove the lemma by induction on $k$. $G\nsg G$ and $G$ contains all 3-cycles, so $G'$ contains all 3-cycles. By the induction hypothesis, $G^{(k)}$ contains all 3-cycles. By the previous lemma, $G^{(k)}\nsg G$, so $G^{(k+1)}=\left(G^{(k)}\right)'$ contains all 3-cycles. $\qed$
Theorem 2.2.10 $S_n$ is not solvable when $n\ge 5$.
Proof. If $S_n$ is solvable then for some $k$, $G^{(k)}=\{e\}$. But also $G^{(k)}$ contains all 3-cycles, a contradiction. $\qed$