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

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 $$n=4$$?

## 1 Answer

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

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.