Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 1D (Groups)

Suppose every group commutator belongs to $H$. For $h\in H$ and $g\in G$, the particular group commutator $h^{-1}{g}^{-1}hg$ belongs to $H$. Multiplying on the left by $h$ gives $g^{-1}hg\in H$. Thus $g^{-1}Hg\subseteq H$ for every $g$; applying the same inclusion to $g^{-1}$ gives the reverse inclusion. Therefore $H$ is a normal subgroup.

We can now form the quotient group. Its group commutators areA group commutator is the identity element exactly when the two elements commute. Hence $G/H$ is an abelian group. Equivalently, the hypothesis says that $H$ contains the commutator subgroup, so the quotient factors through the abelianization.

Conversely, assume $H$ is a normal subgroup and the quotient group $G/H$ is an abelian group. Commuting the two cosets $aH$ and $bH$ givesso every group commutator $a^{-1}{b}^{-1}ab$ belongs to $H$. This proves the converse and completes the equivalence.

For the dihedral group of order ten, choose a rotation $r$ and a reflection $s$ with $r^5=s^2=1$ and $srs=r^{-1}$. Its rotation subgroup $⟨r⟩$ is a normal subgroup, and the quotient by it has order two, hence is an abelian group. Consequently the commutator subgroup is contained in $⟨r⟩$. On the other hand,Since $r^{-2}$ generates the order-five cyclic group $⟨r⟩$, the commutator subgroup is exactly $⟨r⟩$. Every abelian quotient must kill this subgroup. There are only two subgroups containing it, since the remaining quotient has prime order. The complete list, including the trivial quotient, isThis is the order-ten instance of the abelianization of an odd dihedral group.