Abstract Algebra/Composition series

Proof:

We prove the theorem by induction over $$|G|$$.

1. $$|G| = 1$$. In this case, $$G$$ is the trivial group, and $$M_1$$ with $$M_1 = G$$ is a composition series of $$G$$.

2. Assume the theorem is true for all $$n \in \mathbb N$$, $$n < |G|$$.

Since the trivial subgroup $$\{e\} \subset G$$ is a normal subgroup of $$G$$, the set of proper normal subgroups of $$G$$ is not empty. Therefore, we may choose a proper normal subgroup $$N$$ of maximum cardinality. This must also be a maximal proper normal subgroup, since any group in which it is contained must have at least equal cardinality, and thus, if $$M$$ is normal such that
 * $$N \subsetneq M \subsetneq G$$

, then
 * $$|M| > |N|$$

, which is why $$N$$ is not a proper normal subgroup of maximal cardinality.

Due to theorem 2.6.?, $$G / N$$ is simple. Further, since $$|N| < |G|$$, the induction hypothesis implies that there exists a composition series of $$N$$, which we shall denote by $$N_2, \ldots, N_n$$, where
 * $$\{e\} = N_n \triangleleft N_{n-1} \triangleleft \cdots \triangleleft N_2 = N$$

. But then we have
 * $$\{e\} = N_n \triangleleft \cdots \triangleleft N_2 = N \triangleleft N_1 := G$$

, and further for each $$m \in \{1, \ldots, n-1\}$$:
 * $$N_m / N_{m+1}$$ is simple.

Thus, $$N_1, \ldots, N_n$$ is a composition sequence of $$G$$.

Our next goal is to prove that given two normal sequences of a group, we can find two 'refinements' of these normal sequences which are equivalent. Let us first define what we mean by a refinement of a normal sequence.

Proof:

Proof:

Due to theorem 2.6.?, all the elements of $$\{N_1, \ldots, N_n\}$$ must be pairwise different, and the same holds for the elements of $$\{M_1, \ldots, M_k\}$$.

Due to theorem 2.7.4, there exist refinements $$N_1', \ldots, N_m'$$ of $$N_1, \ldots, N_n$$ and $$M_1', \ldots, M_l'$$ of $$M_1, \ldots, M_k$$ such that $$N_1', \ldots, N_m'$$ and $$M_1', \ldots, M_l'$$ are equivalent.

But these refinements satisfy
 * $$\{N_1', \ldots, N_m'\} = \{N_1, \ldots, N_n\}$$

and
 * $$\{M_1', \ldots, M_l'\} = \{M_1, \ldots, M_k\}$$

, since if this were not the case, we would obtain a contradiction to theorem 2.6.?.

We now choose a bijection $$\sigma: \{1, \ldots, m\} \to \{1, \ldots, m\}$$ such that for all $$j \in \{1, \ldots, m-1\}$$:
 * $$N_j/N_{j+1} \cong M_{\sigma(j)}/M_{\sigma(j)+1}$$