Why can this not hold in that case? I am looking at an exercise. 
Let $S_n=H_0\triangleright H_1 \triangleright \ldots \triangleright H_r=\langle 1\rangle$ and $H_{i-1}/H_i$ be abelian. 
Why can the above not hold when each $H_i$ contains all the cycles of length $3$ ?
 A: I am going to assume you meant to say for $n>4$, and for $H_i$ such that $i\neq r$ since this is possible for $n=3$. To see this take $n=3$ $S_3 \rhd A_3 \rhd \{1\}$. The quotients are $\mathbb{Z}_2$ and $\mathbb{Z}_3$ respectively, so the statement holds.  
First think about the normal subgroups of $S_n$ when n<5, these should be easy to classify. Then think about the normal subgroups of $S_n$ when $n \geq 5$. Clearly $A_n$ is a normal subgroup of $S_n$ since $[S_n:A_n]=2$. Now it is a commonly known fact that $A_n$ is simple for $n\geq 5$. You can also show that the only proper nontrivial normal subgroup of $S_n$ is $A_n$ for $n \geq 5$. You can also show that $A_n$ is generated by the set of 3-cycles when $n \geq 3$. To sum it all up:


*

*Show $A_n$ is generated by the set of 3-cycles when $n \geq 3$ 

*Break it up into 2 cases $n=4$ which you can easily prove by exhaustion and then $n\geq 5$. 

*Show $A_n$ is simple for $n\geq 5$ (this should have been stated or proven in most classes, and if it wasn't it's a rather long proof but doable)

*Show that the only proper nontrivial normal subgroup of $S_n$ is $A_n$ for $n\geq 5$.


You should see how it follows from here since $A_n$ is certainly not abelian for $n>3$. 
