## 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.