Inductive step of the Jordan-Holder Theorem

Prove part 2 of the Jordan-Holder Theorem by induction on the shortest length of a composition series. [Hint: apply the inductive hypothesis to N_{k-1} \cap M_{\ell-1} and use the preceding exercises.]


First we prove some lemmas.

Lemma 1: Let G be a group. Let H, K_1, K_2 \leq G be subgroups such that H \vartriangleleft G, K_2 \vartriangleleft K_1, and K_1/K_2 is simple. Then either (H \cap K_1)/(H \cap K_2) \cong K_1/K_2 or H \cap K_1 = K_2. Proof: Note that H \cap K_1 is normal in K_1, so that K_2(H \cap K_1) is normal in K_1. Now K_2 \leq K_2(H \cap K_1) is normal, so by the Lattice Isomorphism Theorem we have (K_2(H \cap K_1))/K_2 \vartriangleleft K_1/K_2. Because K_1/K_2 is simple, there are two possibilities.

  • If K_2(H \cap K_1) = K_1, then since H \cap K_2 \vartriangleleft H \cap K_1 we have
    (H \cap K_1)/(H \cap K_2)  =  (H \cap K_1)/(H \cap K_1 \cap K_2)
    \cong K_2(H \cap K_1)/K_2
     =  K_1/K_2

    by the Second Isomorphism Theorem.

  • If K_2(H \cap K_1) = K_2, then H \cap K_1 \leq K_2. Now if g \in K_2, we have g(H \cap K_1) = gH \cap gK_1 = Hg \cap K_1g = (H \cap K_1)g, so that H \cap K_1 \vartriangleleft K_2. \square

Lemma 2: Let G be a group, H, K_1, K_2 \leq G subgroups such that H \vartriangleleft G, K_2 \vartriangleleft K_1, and K_1/K_2 is simple, and K_2 \leq H. Then either H \cap K_1 = K_1 or H \cap K_1 = K_2. Proof: We have K_2 \leq H \cap K_1 \vartriangleleft K_1. By the Lattice Isomorphism Theorem, (H \cap K_1)/K_2 \vartriangleleft K_1/K_2. The conclusion follows because K_1/K_2 is simple. \square

Lemma 3: Let G be a group, N \vartriangleleft G with G/N simple, and \{ H_i \}_{i=1}^k a family of subgroups of G such that H_i \vartriangleleft H_{i+1} for 1 \leq i < k, H_kN = G, and H_i \not\leq N for all 1 \leq i \leq k. Then H_iN = G for all 1 \leq i \leq k. Proof: We proceed by induction. For the base case, H_kN = G by hypothesis. For the inductive step, suppose 1 \leq i < k such that H_{i+1}N = G. Now N \vartriangleleft H_iN \vartriangleleft H_{i+1}N = G. By the Lattice Isomorphism Theorem, H_iN/N \vartriangleleft G/N. Since H_i \not\leq N, H_iN = G. Thus by induction we have H_iN = G for all 1 \leq i \leq k. \square

Now we move to the main result.

Note that every composition sequence of a finite group is necessarily finite. For a finite group G, we let the length of G be the shortest length of any composition series of G.

Let G be a finite group.

If the length of G is 1, then G is simple and thus has only one composition series; hence all composition series of G have the same length and isomorphic composition factors up to a permutation.

If the length of G is 2, then we saw in the previous exercise that all composition series of G have length 2 and isomorphic composition factors up to a permutation.

Suppose now that for some n > 2, if G is a finite group of length n or less then all composition series of G have the same length and isomorphic composition factors up to a permutation. Suppose G has length n+1, and let (1) \quad 1 = M_0 \vartriangleleft \cdots M_{n+1} = G and (2) \quad 1 = N_0 \vartriangleleft \cdots N_k = G be composition series of G. Let 1 = N_0 \vartriangleleft \cdots \vartriangleleft N_k = G be any other composition series of G; clearly k \geq n+1.

If M_n = N_{k-1}, note that 1 = M_0 \vartriangleleft \cdots \vartriangleleft M_n and 1 = N_0 \vartriangleleft \cdots \vartriangleleft N_{k-1} = M_n are composition series for M_n of length n and k-1, respectively. By the induction hypothesis, we have k = n+1 and a permutation \sigma exists of the set \{0,\ldots,n-1 \} such that M_{i+1}/M_i \cong N_{\sigma(i)+1}/N_{\sigma(i)}. Moreover, G/M_n \cong G/N_n trivially. Thus the composition series (1) and (2) have the same length and are equivalent via the permuation \sigma \cup \{(n,n)\}.

Suppose now that M_n \neq N_{k-1}. If M_n \leq N_{k-1}, then by the Lattice Isomorphism Theorem we have N_{k-1}/M_n \vartriangleleft G/M_n. Because G/M_n is simple, either N_{k-1} = M_n or N_{k-1} = G; either case yields a contradiction. Hence M_n \not\leq N_{k-1}; similarly, N_{k-1} \not\leq M_n.

Note that for all g \in G, we have gM_nN_{k-1}g^{-1} = M_ngg^{-1}N_{k-1} = M_nN_{k-1}, so that M_nN_{k-1} \vartriangleleft G. Because M_n \leq M_nN_{k-1}, by the Lattice Isomorphism Theorem we have M_nN_{k-1}/M_n \vartriangleleft G/M_n. Again because G/M_n is simple, either M_nN_{k-1} = M_n or M_nN_{k-1} = G. In the first case we have N_{k-1} \leq M_n, a contradiction. Thus M_nN_{k-1} = G.

Note that 1 = M_0 \leq N_{k-1}. Let the integer t be maximal such that M_t \leq N_{k-1}; note that 0 \leq t < n. By Lemma 3, we have M_iN_{k-1} = G for t < i \leq n. Consider now the sequence (3) \quad 1 = M_0 \vartriangleleft \cdots \vartriangleleft M_t = M_t \cap N_{k-1} =_\star M_{t+1} \cap N_{k-1} \vartriangleleft \cdots \vartriangleleft M_n \cap N_{k-1}. Note that we have the starred equality by Lemma 2, as M_{t+1} \cap N_{k-1} \not\supseteq M_{t+1} since t is maximal.

  1. If 0 \leq i < t, then M_{i+1}/M_i is simple.
  2. If t+1 \leq i < n, then M_{i+1}N_{k-1} = G. Suppose by way of contradiction that M_{i+1} \cap N_{k-1} \vartriangleleft M_i. Now M_i/(M_{i+1} \cap N_{k-1}) \vartriangleleft M_{i+1}/(M_{i+1} \cap N_{k-1}) \cong M_{i+1}N_{k-1}/N_{k-1} = G/N_{k-1}, by the Second Isomorphism Theorem. Since G/N_{k-1} is simple and M_i \neq M_{i+1}, we have M_i = M_{i+1} \cap N_{k-1}. Then M_i \leq N_{k-1}, a contradiction since t < i. By Lemma 1, then, we have (M_{i+1} \cap N_{k-1})/(M_i \cap N_{k-1}) \cong M_{i+1}/M_i.
  3. In particular, because all intermediate quotients are simple, (3) is a composition series for M_n \cap N_{k-1} of length n-1.

Similarly, if the integer s is maximal such that N_s \leq M_n, then (4) \quad 1 = N_0 \vartriangleleft \cdots \vartriangleleft N_s = M_n \cap N_s = M_n \cap N_{s+1} \vartriangleleft cdots \vartriangleleft M_n \cap N_{k-1} is a composition series for M_n \cap N_{k-1} of length k-2 since for 0 \leq i < s the quotient N_{i+1}/N_i is simple and for s+1 \leq i < k-1 the group (M_n \cap N_{i+1})/(M_n \cap N_i) \cong N_{i+1}/N_i is simple.

By the inductive hypothesis, we have k-2 = n-1, so k = n+1, and there exists a bijection \sigma : \{ 0, \ldots, n-1 \} \setminus \{t\} \rightarrow \{ 0, \ldots, n-1\} \setminus \{s\} such that M_{i+1}/M_i \cong N_{\sigma(i)+1}/N_{\sigma(i)}.

Finally, note that G/N_n = M_{t+1}N_n/N_n \cong M_{t+1}/(M_{t+1} \cap N_n) = M_{t+1}/M_t and G/M_n = M_nN_{s+1}/M_n \cong N_{s+1}/(M_n \cap N_{s+1}) = N_{s+1}/N_s by the Second Isomorphism Theorem. Thus, via the permutation \sigma \cup \{ (t,n), (n,s) \}, the composition series (1) and (2) are equivalent.

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