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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 $3$ ?

  • 2
    Have you tried to see what the subgroup generated by the $3$-cycles looks like? – 2017-01-21
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    Or even simpler: Do all $3$-cycles commute? – 2017-01-21
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    So, when each $H_i$ contains all the cycles of length $3$, are the generated by $3$-cycles? @TobiasKildetoft – 2017-01-21
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    No, they could be larger than that. – 2017-01-21

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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:

  1. Show $A_n$ is generated by the set of 3-cycles when $n \geq 3$
  2. Break it up into 2 cases $n=4$ which you can easily prove by exhaustion and then $n\geq 5$.
  3. 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)
  4. 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 $n>3$.