- #1

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- Homework Statement
- Suppose ##G## has a composition series. Then any normal series of ##G## can be refined into a composition series.

- Relevant Equations
- A normal series of ##G## is

$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$

where ##G_i \trianglelefteq G_{i+1}##. The factors of this normal series are the quotient groups ##G_i / G_{i+1}##. The length of a normal series is the number of distinct factors. A composition series is a normal series whose factors are all simple. Equivalently, a composition series is a normal series of maximal length.

Attempt: Consider an arbitrary normal series ##G = G_0 \ge G_1 \ge G_2 \ge \dots \ge G_n = 1##. We will refine this series into a composition series. We start by adding maximal normal subgroups in between ##G_0## and ##G_1##. If ##G_0/G_1## is simple, then we don't have to do anything. Choose maximal normal subgroup ##G_{0,0}## where ##G_0 \ge G_{0,0} \ge G_1##. Then ##G_0/G_{0,0}## is simple. If ##G_{0.0}/G_1## is also simple, we can stop. Otherwise we keep choosing maximal normal subgroups. Eventually?, we get

$$G = G_0 \ge G_{0,0} \ge G_{0,1} \ge \dots \ge G_{0, k_0} \ge G_1 \ge G_2 \ge \dots \ge G_n = 1$$

where ##G_0/G_{0,0}, G_{0,0}/G_{0,1}, \dots, G_{0, k_0-1}/G_{0, k_0}, G_{0, k_0}/G_1## are simple. But how do we know we eventually find a group ##G_{0, k_0}##? Is it because we're guaranteed a composition series exists?

Can we have a normal series whose length is greater than the length of a composition series?

$$G = G_0 \ge G_{0,0} \ge G_{0,1} \ge \dots \ge G_{0, k_0} \ge G_1 \ge G_2 \ge \dots \ge G_n = 1$$

where ##G_0/G_{0,0}, G_{0,0}/G_{0,1}, \dots, G_{0, k_0-1}/G_{0, k_0}, G_{0, k_0}/G_1## are simple. But how do we know we eventually find a group ##G_{0, k_0}##? Is it because we're guaranteed a composition series exists?

Can we have a normal series whose length is greater than the length of a composition series?