= Solution
Conversely, assume $H$ is a <normal subgroup> and the <quotient group> $G/H$ is an <abelian group>. Commuting the two <cosets> $aH$ and $bH$ gives
$$
(a^{-1}b^{-1}ab)H=H,
$$
so 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> $\langle r\rangle$ is a <normal subgroup>, and the quotient by it has order two, hence is an <abelian group>. Consequently the <commutator subgroup> is contained in $\langle r\rangle$. On the other hand,
$$
[r,s]=r^{-1}s^{-1}rs=r^{-2}.
$$
Since $r^{-2}$ generates the order-five <cyclic group> $\langle r\rangle$, the <commutator subgroup> is exactly $\langle r\rangle$. Every <abelian quotient> must kill this <subgroup>. There are only two <subgroups> containing it, since the remaining quotient has prime order. \b[The complete list, including the trivial quotient, is]
$$
\boxed{D_{10}/\langle r\rangle\cong C_2,\qquad D_{10}/D_{10}\cong\{1\}.}
$$
This is the order-ten instance of the <abelianization of an odd dihedral group>.
Back to article page
