All composition and Chief series for $S_n$ $n \geq 4$

So a composition series is a series $1 \leq G_1 \leq G_2 \leq ... \leq S_n$

s.t. $\frac{G_{i+1}}{G_i}$ is simple.

A Chief series is the same thing except now every factor is characteristically simple i.e. has no characteristic subgroups.

So Is this easy for $n \geq 5$ because then $A_n$ is simple? How do I handle when $n=4$?


$\newcommand{\cyclic}[1]{\langle #1\rangle}%cyclic group$ $\newcommand{\norm}{\triangleleft\,}%normal subgroup$

Just keep factoring stuff out of $S_4$ until you can't anymore!

We can factor out $A_4$ giving us a factor of $\mathbb{Z}_2$. But $A_4$ is not simple. The biggest normal subgroup of it we can find is the one with all the 2+2-cycles in it: $\{1,(12)(34),(13)(24),(14)(23)\}$. Let's factor this out of $A_4$. We get another factor of order 3. But wait, the klein 4-group still isn't simple because it has $\cyclic{(12)(34)}$ as a normal subgroup. Factor that out too.

$$ 1\norm \cyclic{(12)(34)}\norm \{1,(12)(34),(13)(24),(14)(23)\}\norm A_4\norm S_4 $$

Check that all the factors are simple (they are all cyclic groups of order 2 and 3 so we're good), so this is a composition series. The only other one we could get is by taking the two other cyclic groups in the last step.


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