Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 3 / 6D (Algebra and Geometry)

A subgroup $H$ is a normal subgroup of $G$ if $gH{g}^{-1}=H$ for every $g\in G$, equivalently if $gH=Hg$ for every $g$. In the symmetric group $S_3$, the subgroup $A_3=\{e,(123),(132)\}$ is normal: conjugation relabels a three-cycle as a three-cycle, so preserves this set. By contrast, $\{e,(12)\}$ is not normal because conjugating $(12)$ by $(23)$ gives $(13)$ outside the subgroup.

If $H$ is normal, define a product on its left cosets by $(gH)(kH)=(gk)H$. To check that this quotient group operation is well-defined, take different representatives $g{h}_1$ and $k{h}_2$, with $h_1,h_2\in H$. Their product isNormality puts the last two factors in $H$, so its coset is exactly $gkH$. Associativity follows from that in $G$, the identity coset is $H$, and the inverse of $gH$ is $g^{-1}H$. Thus all the group axioms hold, and the projection $g\mapsto gH$ is a group homomorphism.

Always true. Suppose $G=⟨g⟩$. If $H$ is nontrivial, let $d$ be the least positive integer with $g^d\in H$. For any $g^k\in H$, divide $k=qd+r$ with $0\le r<d$. Since $g^r=g^k(g^d{)}^{-q}\in H$, minimality forces $r=0$. Hence $H=⟨g^d⟩$ is a cyclic group. The trivial subgroup is cyclic as well. Every coset in $G/H$ is $g^{k}H=(gH{)}^k$, so the quotient group is generated by $gH$. This proves the subgroups and quotients of a cyclic group assertion.

Can be false. Take the Klein four-group $G=C_2\times C_2$ and $H=C_2\times \{e\}$. It is a normal subgroup because $G$ is an abelian group. Both $H$ and $G/H$ have order two and are cyclic groups. But every nonidentity element of $G$ has order two, so none generates its four elements. Thus $G$ is not cyclic.

Always true. Elements of a subgroup $H$ commute when they are elements of an abelian group $G$, so $H$ is abelian. In the quotient group, $(gH)(kH)=gkH=kgH=(kH)(gH)$. Therefore $G/H$ is abelian too.

Can be false. Use $G=S_3$ and the normal subgroup $H=A_3$ from the root solution. The subgroup $H$ is cyclic of order three, hence an abelian group, and the quotient group has the two cosets $A_3$ and $(12)A_3$, so is cyclic of order two and abelian. Yet $(12)(23)=(123)$ whereas $(23)(12)=(132)$, proving that $S_3$ is not abelian. An abelian kernel and an abelian quotient do not force all elements of the original group to commute.