Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 3 / 6D (Groups)

Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.

Writing $D_8=⟨r,s:r^4=s^2=e,srs=r^{-1}⟩$, one has $[r,s]=r^{-2}=r^2$. The quotient by $⟨r^2⟩$ is $C_2\times C_2$, so

The subgroup $A_5^{\prime }$ is normal in $A_5$. It is nontrivial because $A_5$ is nonabelian; simplicity of $A_5$ therefore gives

The sign map has abelian quotient, so $S_5^{\prime }\le A_5$. On the other hand $A_5^{\prime }=A_5$ and $A_5\le S_5$ imply $A_5\le S_5^{\prime }$. Hence