Problem 5.4.1 ($[A,B] = [B,A]$)

Question

Prove that if $x,y \in G$, then $[y,x] = [x,y]^{-1}$. Deduce that for any subsets $A$ and $B$ of $G$, $[A,B] = [B,A]$ (recall that $[A,B]$ is the {\bf subgroup} of $G$ generated by the commutators $[a,b]$).

Richard Ganaye · Accepted Answer

\begin{proof} 
If $x,y \in G$, then
$$[y,x] = y^{-1} x^{-1} y x = (x^{-1} y^{-1} x y)^{-1}= [x,y]^{-1}.$$
If $a \in A$ and $b \in B$, then $[a,b] = [b,a]^{-1} \in [B,A]$.

Let $S$ be the set $S = \{[a,b] \mid a \in A, \ b \in B\}$. Then $S \subseteq [B,A]$ since $[a,b] \in [B,A]$ for all $[a,b] \in S$. Since $[A,B]$ is by definition the smallest subgroup of $G$ which contains $S$, we obtain $[A,B] \subseteq [B,A]$. Exchanging the roles of $A$ and $B$, we obtain similarly, $[B,A] \subseteq [A,B]$, so
$$[A,B] = [B,A].$$
\end{proof}

Problem 5.4.1 ($[A,B] = [B,A]$)

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Problem 5.4.1 ($[A,B] = [B,A]$)

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