Problem 5.4.19 (Perfect groups)

Proof. A group $H$ is called perfect if $H^{\prime }=H$.

(a)

Let $H$ be a non abelian simple group. Then $H^{\prime }⊴H$ and $H^{\prime }\ne \{1\}$, otherwise $H$ is abelian. Since $H$ is simple, $H^{\prime }=H$, so $H$ is perfect.

(b)

Suppose that $H$ and $K$ are perfect subgroups of $G$. Put $L=⟨H,K⟩=⟨H\cup K⟩$. Since $H$ and $K$ are subgroups of $G$, every element $x$ of $L$ is of the form $$x=a_1{a}_2\cdots a_k,\text{where for every }i\text{, }{a}_i\in H\text{ or }{a}_i\in K.$$

Since $H^{\prime }=H$, if $a_i\in H$, then $a_i={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j_i\in I_i}[b_{j_i},c_{j_i}]$ is a (finite) product of commutators of elements $b_{j_i},c_{j_i}\in H$, and similarly if $a_i\in K$, then $a_i={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j_i\in I_i}[b_{j_i},c_{j_i}]$ is a product of commutators of elements of $K$. Then

$$x=\underset{j_1\in I_1}{\prod }[b_{j_{i_1}},c_{j_{i_1}}]\cdots \underset{j_k\in I_k}{\prod }[b_{j_{i_k}},c_{j_{i_k}}],\text{where }{b}_{j_{i_l}},c_{i_{i_l}}\in H\cup K.$$

So $x$ is in the subgroup generated by the commutators of $L=H\cup K$, i.e., $x\in {⟨H\cup K⟩}^{\prime }=L^{\prime }$. This shows

$$L\le L^{\prime }.$$

Since $L^{\prime }\le L$ is always true, we obtain $L=L^{\prime }$, so $L=⟨H,K⟩$ is a perfect subgroup.

More generally, let ${(H_i)}_{i\in I}$ be any family of perfect subgroups of $G$ (where $I$ is finite or not), and put $L=⟨{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{H}_i⟩$. Then every element of $L$ is of the form

$$x=a_1{a}_2\dots a_k,\text{where for every }i\text{, }{a}_i\in \underset{i\in I}{\bigcup }{H}_i.$$

Since $H_j^{\prime }=H_j$ for all $j$, if $a_i\in H_j$, then $a_i=[b_j,c_j]$ is a commutator of elements $b_j,c_j\in H_j$. Then $x$ is a product of commutators $[b,c]$ where $b,c\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{H}_i$, so $x\in L^{\prime }$. As above, this gives $L=L^{\prime }$, so $L=⟨{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{H}_i⟩$ is a perfect subgroup of $G$.

(c)

Suppose that $H=H^{\prime }$, where $H$ is a subgroup of $G$. Consider the conjugate $K=\mathit{𝑔𝐻}{g}^{-1}$, where $g\in G$.

If $x\in K$, then $x=\mathit{𝑔h}{g}^{-1}$, where $h\in H$. Since $H^{\prime }=H$, $h={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}[b_i,c_i]$ is a product of commutators of element of $H$, thus

$$x=\mathit{𝑔h}{g}^{-1}=g\underset{i\in I}{\prod }[b_i,c_i]g^{-1}=\underset{i\in I}{\prod }[g{b}_i{g}^{-1},g{c}_i{g}^{-1}]\in K^{\prime },$$

since $g{b}_i{g}^{-1},g{c}_i{g}^{-1}\in K$ for all $i$. Therefore $K\le K^{\prime }$, where $K^{\prime }\le K$, thus $K=K^{\prime }$.

Any conjugate of a perfect subgroup is perfect.

(d)

Consider the set $S$ of all perfect subgroups of $G$: $$S=\{H\le G\mid H^{\prime }=H\}.$$

Note that $S\ne \varnothing$, because the trivial subgroup $\{1\}$ is in $S$.

By part (b),

$$L=⟨\underset{H\in S}{\bigcup }H⟩$$

is perfect, and contains all perfect subgroups of $G$, so $L$ is a maximal perfect subgroup of $G$. If $L_1$ and $L_2$ are bot maximal perfect subgroups of $G$, then $L_1\le L_2$ and $L_2\le L_1$, so $L_1=L_2$. Thus $L$ is the unique maximal perfect subgroup of $G$.

If $g\in G$, then $\mathit{𝑔𝐿}{g}^{-1}$ is perfect by part (c). Since $L$ contains all perfect subgroups, then $\mathit{𝑔𝐿}{g}^{-1}\le L$. This proves $L⊴G$.