Show that if $K$ is a normal subgroup of $H$ such that $H/K$ is abelian, then K contains all of the 3-cycles.

Question

Claim: For $n≥5$, if $H$ is a subgroup of $S_n$ which contains all of the 3-cycles, and $K$ is a normal subgroup of $H$ such that $H/K$ is abelian, then K also contains all of the 3-cycles.

Attempt: let $f:H\to H/K$. I've already shown that for a group $G$ and $G'$ the subgroup generated by $\{ghg^{-1}h^{-1} \; : \; g,h \in G \}$; if $f:G\to H$ is a homomorphism then $Im(f)$ is abelian iff $G' ⊂ ker(f)$. How do I use this to show that $H' ⊂ K$? How do I show that every 3-cycle belongs to $H'$? (i.e express an arbitrary 3-cycle $(a_1a_2a_3)$ as $ghg^{-1}h^{-1}$ for some 3-cycles $g,h$ so that $g,h \in H$ and $(a_1a_2a_3) \in $H'$.)

Answer 1

You have $f:H \to H/K$ defined by $f(h)=hK$ is a homomorphism (the natural projection). Since $H/K$ is abelian, the image of $f$ is abelian and thus $H' \subset ker(f)$. But $ker(f)=K$ (because $f(h)=hK=K$ iff $h \in K$).

Now consider a 3-cycle $(a_1a_2a_3)$. Since $n \geq 5$, there are some numbers $a_4$ and $a_5$ disjoint from $a_1,a_2,a_3$. Note that if we let $x=(a_1a_2a_4),y=(a_1a_3a_5)$. Then since $H$ contains all 3-cycles $x,y \in H$. Therefore, $$xyx^{-1}y^{-1} = (a_1a_2a_4)(a_1a_3a_5)(a_1a_4a_2)(a_1a_5a_3)=(a_1a_2a_3) \in H'$$

Now $H' \subseteq K$. So $K$ contains all 3-cycles.