# The Jordan-Hölder theorem

by cup
Tags: jordanhölder, theorem
 P: 27 Hello friends. I am working trough "Abstract Algebra" by Dummit & Foote. I recently got to section 3.4, on composition series and "the Hölder program". The Jordan-Hölder theorem states: Let G be a finite, non-trivial group. Then: 1) G has a composition series. 2) If $$\{ 1 \} = N_0 \leq N_1 \leq ... \leq N_r = G$$ and $$\{ 1 \} = M_0 \leq M_1 \leq ... \leq M_s = G$$ are two composition series of G, then: 2a) r = s 2b) There is some permutation $$\pi$$ of {1, 2, ..., r} such that: $$M_{\pi (i)} / M_{\pi (i) - 1} \cong N_{i} / N_{i-1}$$ for $$0 \leq i \leq r$$. My question is on 2b). For $$M_{\pi (i)} / M_{\pi (i) - 1}$$ to make sense, we must of course have $$M_{\pi (0)} \leq M_{\pi (1)} \leq ... \leq M_{\pi (s)}$$ But how can any permutation satisfy this relation, i.e. not break the subgroup ordering? Let me rephrase my question with numbers instead of subgroups. It is clearly impossible to permute the sequence: $$1 \leq 3 \leq 4 \leq 7 \leq 9 \leq 11$$ without breaking the ordering. The only example I can think of where we may permute is something like: $$1 \leq 3 \leq 4 \leq 7 \leq 7 \leq 7$$ Where we are allowed to permute only the last three numbers without breaking the ordering. But the analog of this last example for subgroups would be pretty pointless, since the assertion of 2b would of course hold true before applying such a permutation. Can somebody please explain what I'm missing? Edit: Hmm... Is there perhaps one permutation $$\pi_i$$, say, for each quotient $$N_{i} / N_{i-1}$$, instead of just a single one for every quotient?
 P: 27 Never mind. I found my own error. I was thinking $$M_{\pi (i)} / M_{\pi (i-1)}$$ instead of $$M_{\pi (i)} / M_{\pi (i) - 1}$$ which is what the theorem says. Sorry about that. Cool theorem, though.