1. **Problem statement:** We have a group $G$ and a subgroup $G'$ generated by all elements of the form $aba^{-1}b^{-1}$ for $a,b \in G$. We want to prove: (a) $G'$ is a normal subgroup of $G$. (b) The quotient group $G/G'$ is abelian. (c) For any normal subgroup $H$ of $G$, $G/H$ is abelian if and only if $G' \leq H$. 2. **Recall definitions and facts:** - The element $aba^{-1}b^{-1}$ is called a commutator. - $G'$ is called the commutator subgroup or derived subgroup. - A subgroup $N$ is normal ($N \triangleleft G$) if $gNg^{-1} = N$ for all $g \in G$. - The quotient group $G/H$ is abelian if for all $a,b \in G$, $(aH)(bH) = (bH)(aH)$. 3. **(a) Prove $G' \triangleleft G$:** - Since $G'$ is generated by all commutators, it suffices to show that conjugating a commutator by any $g \in G$ yields another commutator. - Take any commutator $c = aba^{-1}b^{-1}$. - Consider conjugation by $g$: $$g c g^{-1} = g (aba^{-1}b^{-1}) g^{-1} = (g a g^{-1})(g b g^{-1})(g a g^{-1})^{-1}(g b g^{-1})^{-1}$$ - Since $g a g^{-1}, g b g^{-1} \in G$, $g c g^{-1}$ is also a commutator. - Because $G'$ is generated by all commutators, conjugation by $g$ sends generators of $G'$ to elements of $G'$, so $G'$ is normal. 4. **(b) Prove $G/G'$ is abelian:** - In $G/G'$, elements are cosets $aG'$. - Compute the product: $$(aG')(bG') = (ab)G'$$ - Similarly, $$(bG')(aG') = (ba)G'$$ - To show $G/G'$ is abelian, we want $(ab)G' = (ba)G'$, or equivalently, $$ab(ba)^{-1} \in G'$$ - Note that $$ab(ba)^{-1} = ab a^{-1} b^{-1}$$ - This is exactly a commutator, so it lies in $G'$. - Hence, multiplication in $G/G'$ is commutative. 5. **(c) Prove $G/H$ abelian iff $G' \leq H$ for $H \triangleleft G$:** - ($\Rightarrow$) Suppose $G/H$ is abelian. - Then for all $a,b \in G$, $$(aH)(bH) = (bH)(aH) \implies abH = baH \implies ab(ba)^{-1} \in H$$ - Since $ab(ba)^{-1}$ is a commutator, all commutators lie in $H$. - Because $G'$ is generated by all commutators, $G' \leq H$. - ($\Leftarrow$) Suppose $G' \leq H$. - Then for all $a,b \in G$, $$ab(ba)^{-1} \in G' \leq H \implies abH = baH$$ - So multiplication in $G/H$ is commutative, i.e., $G/H$ is abelian. **Final answers:** (a) $G'$ is normal in $G$. (b) $G/G'$ is abelian. (c) $G/H$ is abelian if and only if $G' \leq H$.