Jordan-Hölder Uniqueness

Any two composition series of a group have the same simple composition factors up to order.

Jordan-Hölder Uniqueness

Jordan-Hölder Uniqueness: Let $G$ be a group that admits a composition series . Suppose

(1)=G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G

and

(1)=H_0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_m = G

are composition series (so each successive quotient group $G_i/G_{i-1}$ and $H_j/H_{j-1}$ is a simple group ). Then:

$n=m$ (the lengths coincide), and
there exists a permutation $\sigma$ of $\{1,\dots,n\}$ such that $G_i/G_{i-1} \cong H_{\sigma(i)}/H_{\sigma(i)-1}\quad\text{for all }i.$

This is precisely the “uniqueness” conclusion extracted from the Jordan-Hölder theorem ; a standard proof proceeds via Schreier refinement , showing that any two subnormal series have equivalent refinements.

Examples:

In $S_3$ , the chain $(1)\triangleleft A_3 \triangleleft S_3$ is a composition series; the factors are $A_3/(1)\cong C_3$ and $S_3/A_3\cong C_2$ . Any other composition series of $S_3$ yields the same multiset $\{C_2,C_3\}$ .
In the cyclic group $C_{12}$ , one composition series is $(1)\triangleleft C_2 \triangleleft C_4 \triangleleft C_{12}$ with factors $C_2, C_2, C_3$ . Another is $(1)\triangleleft C_3 \triangleleft C_6 \triangleleft C_{12}$ with factors $C_3, C_2, C_2$ . The factors match up to reordering, as the theorem predicts.