----- > [!proposition] Proposition. ([[universal property of commutator subgroup]]) > > Let $G'$ denote the [[commutator subgroup]] of a [[group]] $G$. > > 1. $G / G'$ is [[abelian group|abelian]]. More generally, $G / N$ is [[abelian group|abelian]] for any $N \geq G'$. > 2. If $\psi \in \hom(G,A)$ where $A$ is [[abelian group|abelian]], then $G' \leq \ker \psi$ and there is a unique [[group homomorphism|homomorphism]] $\overline{\psi}:G / G' \to A$ such that the following diagram commutes (where $\pi:G \to G / G'$ is the [[kernel iff normal subgroup|NPH]]): > ```tikz > \usepackage{tikz-cd} > \begin{document} > \begin{tikzcd} > G \arrow{d}{\pi} \arrow{r}{\forall \psi} & A \\ > G/G' \arrow{ur}{\exists ! \bar{\psi}} > \end{tikzcd} > \end{document} > ``` > > 3. The [[commutator subgroup]] is the $\subset$-smallest $H \trianglelefteq G$ such that $G / H$ is [[abelian group|abelian]]. That is, if $G / N$ is [[abelian group|abelian]] for some $N \trianglelefteq G$, then $H=G' \leq N$. > > *In summary*: the [[quotient by N is abelian iff N contains the commutator subgroup]]. > [!proof]- Proof. ([[universal property of commutator subgroup]]) > ~ **1.** Let $\overline{x} :=G'x, \overline{y}:=G'y \in G / G'$. Then simply $\begin{align} \overline{x} \ \overline{y} = & (G'x)(G'y) \\ = & G' xy \\ = & G'\textcolor{Thistle}{yxy^{-1}x ^{-1}} x y \ (G' \text{ contains } [y,x] ) \\ = & G' y x \\ = & (G'y)(G'x) \\ = & \overline{y} \ \overline{ x}. \text{ } \end{align}$ (Or $G'=G' x y x ^{-1} y ^{-1}$ and rearrange) More generally, if $N \geq G'$ with $\overline{x}:=Nx, \overline{y} := Ny\in N$ then the same proof goes through because $N$ contains $[y,x]$. **2.** $[G,G]$ is [[generating set of a group|generated by]] [[commutator|commutators]], so it suffices to show that $\psi: G \to A$ kills any [[commutator]] $xyx ^{-1} y ^{-1}$. Because $A$ is [[abelian group|abelian]]: $\psi(xyx ^{-1} y ^{-1})=\psi(x)\psi(y)\psi(x)^{-1} \psi(y)^{-1}=\psi(x)\psi(x)^{-1}\psi(y)\psi(y)^{-1}=e_{A}.$ So $[G,G] \subset \ker \psi$. The result now follows from the [[characterization of quotienting a group]]. **3.** Consider some $H \trianglelefteq G$ for which $G / H$ is [[abelian group|abelian]]. Then for all $x,y \in G$ we get $H[x,y]=Hx y x ^{-1} y ^{-1}=H(xx ^{-1})(y y^{-1})=H,$ hence $[x,y] \in H$. Since every element of $[G,G]$ is a word in [[commutator|commutators]] and $H$ is closed under multiplication it follows that $[G,G] \leq H$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```