[proofplan] The proof has two independent halves: existence and uniqueness. **Existence** follows by strong induction on $|G|$: pick a maximal proper normal subgroup $N \triangleleft G$ (which exists because $G$ is finite), inductively obtain a composition series of $N$, and prepend $G$ to it; the new top quotient $G/N$ is simple by maximality of $N$. **Uniqueness** is the heart of the theorem and follows from the Zassenhaus butterfly lemma, which shows that any two normal series of $G$ admit a common refinement with isomorphic factors (Schreier's refinement theorem). Applied to two composition series — which by simplicity admit no proper refinements — this forces the two series themselves to have the same factors up to permutation and isomorphism. [/proofplan] [step:Set up the existence proof by strong induction on $|G|$] We prove existence of a composition series by strong induction on $n := |G| \geq 2$. **Base case** ($n = 2$): then $G \cong \mathbb{Z}/2\mathbb{Z}$ is simple, so the series $\{e\} \triangleleft G$ is itself a composition series with the single simple factor $G/\{e\} \cong G$. **Inductive step**: suppose $n \geq 3$ and the result holds for all groups of order strictly less than $n$. [guided] We argue by strong induction on the order $|G|$. The base case is the smallest non-trivial finite group: $|G| = 2$ forces $G \cong \mathbb{Z}/2\mathbb{Z}$, which has no proper non-trivial subgroup at all and is therefore simple. The one-step series $\{e\} \triangleleft G$ has the one factor $G/\{e\} \cong G$, which is simple, so this is a composition series. For the inductive step, fix $n \geq 3$ and assume that every group of order strictly less than $n$ admits a composition series. We must construct a composition series for our given $G$ of order $n$. The strategy is to find a "largest possible" proper normal subgroup $N$ — this will guarantee that $G/N$ is simple — and then attach $G$ to a composition series for $N$ obtained by induction. [/guided] [/step] [step:Choose a maximal proper normal subgroup $N$ of $G$ using finiteness] Let $\mathcal{N} := \{N \triangleleft G : N \neq G\}$ be the set of proper normal subgroups of $G$. Then $\{e\} \in \mathcal{N}$, so $\mathcal{N}$ is non-empty. Since $G$ is finite, $\mathcal{N}$ is a finite set, so it contains an element $N$ of maximal order. By construction, $N \triangleleft G$ and $N \neq G$. [guided] We need to choose a "largest possible" proper normal subgroup. The collection \begin{align*} \mathcal{N} := \{N \triangleleft G : N \neq G\} \end{align*} is non-empty because the trivial subgroup $\{e\}$ is normal in any group and is proper since $|G| \geq 3 > 1$. Because $G$ is finite, $\mathcal{N}$ is a finite collection of subsets of $G$, so the orders $\{|N| : N \in \mathcal{N}\}$ form a finite subset of $\mathbb{N}$ and therefore attain a maximum. Pick $N \in \mathcal{N}$ with $|N|$ maximal. Why does this matter? The crucial property of $N$ is that there is no normal subgroup of $G$ strictly between $N$ and $G$. By the [Lattice Isomorphism Theorem](/theorems/854), normal subgroups of $G/N$ correspond bijectively to normal subgroups of $G$ that contain $N$. So the absence of an intermediate normal subgroup will translate into simplicity of $G/N$ — which is exactly what we need for the top of a composition series. [/guided] [/step] [step:Verify that $G/N$ is simple via the Lattice Isomorphism Theorem] [claim:$G/N$ is simple] [proof] By the [Lattice Isomorphism Theorem](/theorems/854) (a.k.a. the Correspondence Theorem) applied to the quotient map $\pi: G \to G/N$, the map \begin{align*} \Phi: \{H : N \leq H \leq G\} &\to \{\overline{H} : \overline{H} \leq G/N\} \\ H &\mapsto H/N \end{align*} is an inclusion-preserving bijection that restricts to a bijection between normal subgroups of $G$ containing $N$ and normal subgroups of $G/N$. Suppose $\overline{M} \triangleleft G/N$. Then $\overline{M} = M/N$ for a unique $M$ with $N \leq M \leq G$, and $M \triangleleft G$. Since $N \leq M$ and $N$ was chosen to have maximal order among proper normal subgroups, either $M = N$ (giving $\overline{M} = \{e\}$) or $M = G$ (giving $\overline{M} = G/N$). Hence $G/N$ has no normal subgroups other than $\{e\}$ and $G/N$ itself. Finally, $G/N$ is non-trivial because $N \neq G$. So $G/N$ is simple. [/proof] [/claim] [guided] The claim is that $G/N$ has exactly two normal subgroups: the identity subgroup and itself. We use the **Lattice Isomorphism Theorem** for the quotient map $\pi: G \twoheadrightarrow G/N$. This theorem states that the assignment \begin{align*} \Phi: \{H \leq G : N \leq H\} &\to \{\overline{H} \leq G/N\} \\ H &\mapsto H/N = \pi(H) \end{align*} is an inclusion-preserving bijection, and that $\Phi$ restricts to a bijection on the corresponding sublattices of normal subgroups: $H \triangleleft G$ (with $H \supseteq N$) if and only if $H/N \triangleleft G/N$. Now suppose $\overline{M} \triangleleft G/N$. By the bijection, $\overline{M} = M/N$ for a uniquely determined $M$ with $N \leq M \leq G$ and $M \triangleleft G$. We chose $N$ to have **maximal** order among proper normal subgroups of $G$, so any $M \triangleleft G$ with $M \supseteq N$ satisfies $|M| \geq |N|$, and equality would force $M = N$ (since $M \supseteq N$); strict inequality forces $M \notin \mathcal{N}$, i.e., $M = G$. So $\overline{M} \in \{N/N, G/N\} = \{\{e\}, G/N\}$. Together with $G/N \neq \{e\}$ (because $N$ is a *proper* subgroup), this means $G/N$ is simple. [/guided] [/step] [step:Apply the inductive hypothesis to $N$ and prepend $G$] If $N = \{e\}$, then by the previous step $G \cong G/\{e\}$ is simple, and $\{e\} \triangleleft G$ is a composition series. Otherwise $1 < |N| < |G| = n$, so the inductive hypothesis yields a composition series \begin{align*} \{e\} = N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_{k-1} = N \end{align*} with each $N_i \triangleleft N_{i+1}$ and each factor $N_{i+1}/N_i$ simple. Setting $N_k := G$, we obtain \begin{align*} \{e\} = N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_{k-1} = N \triangleleft N_k = G, \end{align*} in which $N_{k-1} = N \triangleleft G = N_k$ holds by our choice of $N$, and the new top factor $N_k/N_{k-1} = G/N$ is simple by the previous step. All other factors are simple by the inductive hypothesis. This is a composition series for $G$, completing the existence proof. [guided] With $N$ in hand, we have two cases for finishing the induction. **Case 1: $N = \{e\}$.** Then the previous step gives that $G/\{e\} \cong G$ is simple. The one-step series $\{e\} \triangleleft G$ is therefore a composition series with the one simple factor $G$. **Case 2: $N \neq \{e\}$.** Then $1 < |N| < n$, since $N$ is a *proper* subgroup. The strict inequality $|N| < n$ is exactly what we need to invoke the **inductive hypothesis** on $N$, yielding a composition series \begin{align*} \{e\} = N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_{k-1} = N, \end{align*} where each $N_i \triangleleft N_{i+1}$ inside $N$, and each successive quotient $N_{i+1}/N_i$ is simple. Now we extend this series to all of $G$ by setting $N_k := G$. We must check two things to verify this is a composition series for $G$. First, is $N_{k-1} = N$ normal in $N_k = G$? Yes — by construction, $N \triangleleft G$. Second, are all factors simple? The factors $N_{i+1}/N_i$ for $i < k - 1$ are simple by the inductive hypothesis applied to $N$. The new top factor is $N_k/N_{k-1} = G/N$, which we proved simple in the previous step. Both conditions hold, so $\{e\} = N_0 \triangleleft N_1 \triangleleft \cdots \triangleleft N_{k-1} \triangleleft N_k = G$ is a composition series. Existence is established. [/guided] [/step] [step:State the Zassenhaus butterfly lemma needed for uniqueness] For uniqueness we need the following technical lemma. Recall that for subgroups $A, B$ of a group, the join $\langle A, B \rangle$ is the smallest subgroup containing both, and we write $AB$ for $\{ab : a \in A, b \in B\}$ (which equals $\langle A, B \rangle$ when $A$ normalises $B$ or vice versa). [claim:Zassenhaus's butterfly lemma] Let $A_0 \triangleleft A$ and $B_0 \triangleleft B$ be subgroups of a group $G$. Then $A_0(A \cap B_0) \triangleleft A_0(A \cap B)$ and $B_0(A_0 \cap B) \triangleleft B_0(A \cap B)$, and there is an isomorphism \begin{align*} \frac{A_0(A \cap B)}{A_0(A \cap B_0)} \cong \frac{B_0(A \cap B)}{B_0(A_0 \cap B)}. \end{align*} [proof] Both quotients are isomorphic to the common quotient \begin{align*} \frac{A \cap B}{(A_0 \cap B)(A \cap B_0)}, \end{align*} via the second isomorphism theorem applied symmetrically in $A$ and $B$. The full verification — checking normality of all four subgroups appearing and constructing the two isomorphisms via the [Second Isomorphism Theorem](/theorems/843) — is standard; see for instance Lang, *Algebra*, Chapter I §3, or Zassenhaus's Lemma for the detailed proof. [/proof] [/claim] [guided] Zassenhaus's lemma is the technical engine of the uniqueness argument. The setup is symmetric in $(A_0, A)$ and $(B_0, B)$: we have two pairs of "bigger normal in smaller normal" subgroups, and the lemma produces a canonical isomorphism between two quotients that mix them. To remember the statement, picture a Hasse diagram with $A$ above $A_0$ on the left, $B$ above $B_0$ on the right, and the four "mixed" subgroups $A_0(A \cap B)$, $A_0(A \cap B_0)$, $B_0(A \cap B)$, $B_0(A_0 \cap B)$ arranged in the middle — the picture is the "butterfly" that gives the lemma its name. The proof shows that both of the quotients in the conclusion are isomorphic to a third, perfectly symmetric, quotient \begin{align*} \frac{A \cap B}{(A_0 \cap B)(A \cap B_0)}. \end{align*} This common middle term gives the isomorphism. The verification that $(A_0 \cap B)(A \cap B_0)$ is a normal subgroup of $A \cap B$ and that the relevant maps factor through it uses the [Second Isomorphism Theorem](/theorems/843) twice — once for each side of the butterfly. We take this as known; the full bookkeeping is in any standard algebra text (Lang's *Algebra*, Hungerford's *Algebra*, or Rotman's *An Introduction to the Theory of Groups*). [/guided] [/step] [step:Derive Schreier's refinement theorem from Zassenhaus] Recall: a **subnormal series** of $G$ is a chain $G = G_0 \triangleright G_1 \triangleright \cdots \triangleright G_r = \{e\}$ with each $G_{i+1} \triangleleft G_i$. A **refinement** of such a series inserts more terms between existing ones (preserving subnormality). Two subnormal series are **equivalent** if they have the same length and their factors agree up to permutation and isomorphism. [claim:Schreier's refinement theorem] Any two subnormal series of $G$ have equivalent refinements. [proof] Let \begin{align*} G &= G_0 \triangleright G_1 \triangleright \cdots \triangleright G_r = \{e\}, \\ G &= H_0 \triangleright H_1 \triangleright \cdots \triangleright H_s = \{e\} \end{align*} be two subnormal series. For each $i \in \{0, \ldots, r-1\}$ and $j \in \{0, \ldots, s\}$, define \begin{align*} G_{i,j} := G_{i+1}(G_i \cap H_j). \end{align*} Then $G_{i,0} = G_{i+1}(G_i \cap G) = G_i$ and $G_{i,s} = G_{i+1}(G_i \cap \{e\}) = G_{i+1}$, so the chain \begin{align*} G_i = G_{i,0} \supseteq G_{i,1} \supseteq \cdots \supseteq G_{i,s} = G_{i+1} \end{align*} inserts $s$ subgroups between $G_i$ and $G_{i+1}$. Concatenating these refined chunks for $i = 0, 1, \ldots, r-1$ produces a refinement of the first series of length $rs$. Symmetrically, define \begin{align*} H_{j,i} := H_{j+1}(H_j \cap G_i) \end{align*} for $j \in \{0, \ldots, s-1\}$, $i \in \{0, \ldots, r\}$, producing a refinement of the second series, also of length $rs$. By Zassenhaus's butterfly lemma applied with $A_0 = G_{i+1}$, $A = G_i$, $B_0 = H_{j+1}$, $B = H_j$, we obtain \begin{align*} \frac{G_{i,j}}{G_{i,j+1}} = \frac{G_{i+1}(G_i \cap H_j)}{G_{i+1}(G_i \cap H_{j+1})} \cong \frac{H_{j+1}(H_j \cap G_i)}{H_{j+1}(H_j \cap G_{i+1})} = \frac{H_{j,i}}{H_{j,i+1}}. \end{align*} This pairs the $rs$ factors of the first refinement bijectively with the $rs$ factors of the second refinement, with corresponding factors isomorphic. Hence the two refinements are equivalent. [/proof] [/claim] [guided] Schreier's theorem says: any two subnormal series have a common refinement, in the strong sense that not only do they refine to series of the same length, but the *factors* of those refined series can be matched up so corresponding pairs are isomorphic. The construction is symmetric and combinatorial. Given two series of lengths $r$ and $s$, we form a "doubly indexed" refinement by inserting $s - 1$ intermediate subgroups between each consecutive pair $G_i \triangleright G_{i+1}$ of the first series, using the second series as a probe: \begin{align*} G_{i,j} := G_{i+1}(G_i \cap H_j), \quad j = 0, 1, \ldots, s. \end{align*} At $j = 0$ we get $G_i$ (since $H_0 = G$ so $G_i \cap H_0 = G_i \supseteq G_{i+1}$); at $j = s$ we get $G_{i+1}$ (since $H_s = \{e\}$ so $G_i \cap H_s = \{e\}$). In between, we get a chain that descends from $G_i$ to $G_{i+1}$. Symmetrically, using the first series to probe the second, we get $H_{j,i} := H_{j+1}(H_j \cap G_i)$. The refined series both have length $rs$. The Zassenhaus butterfly lemma — applied with the substitution $A_0 = G_{i+1}$, $A = G_i$, $B_0 = H_{j+1}$, $B = H_j$ — provides the isomorphism between the $(i,j)$-th factor of one refinement and the $(j,i)$-th factor of the other: \begin{align*} \frac{G_{i+1}(G_i \cap H_j)}{G_{i+1}(G_i \cap H_{j+1})} \cong \frac{H_{j+1}(H_j \cap G_i)}{H_{j+1}(H_j \cap G_{i+1})}. \end{align*} The bijection $(i, j) \leftrightarrow (j, i)$ between index sets is a permutation of the $rs$ factors; corresponding factors are isomorphic. Hence the refinements are equivalent. [/guided] [/step] [step:Conclude uniqueness by observing composition series admit no proper refinement] Let \begin{align*} G &= G_0 \triangleright G_1 \triangleright \cdots \triangleright G_k = \{e\}, \\ G &= H_0 \triangleright H_1 \triangleright \cdots \triangleright H_\ell = \{e\} \end{align*} be two composition series. By Schreier's refinement theorem they have equivalent refinements. We now use simplicity to argue that the refinements must equal the original series, modulo repeated terms. A refinement inserts subgroups; a composition series has all factors $G_i/G_{i+1}$ simple. Inserting a subgroup $G_{i+1} \trianglelefteq K \trianglelefteq G_i$ produces factors $G_i/K$ and $K/G_{i+1}$. By the [Lattice Isomorphism Theorem](/theorems/854), $K/G_{i+1}$ corresponds to a normal subgroup of the simple group $G_i/G_{i+1}$, hence equals either the trivial subgroup (so $K = G_{i+1}$) or all of $G_i/G_{i+1}$ (so $K = G_i$). In either case the inserted subgroup duplicates an existing term, producing a factor isomorphic to $\{e\}$. Removing $\{e\}$-factors from a refinement returns the original series. Thus, after removing $\{e\}$-factors, the two equivalent refinements reduce to the original two composition series. Since the refinements have equivalent (i.e., bijective up to isomorphism) lists of factors, and removing $\{e\}$-factors from each produces the original lists $(G_0/G_1, \ldots, G_{k-1}/G_k)$ and $(H_0/H_1, \ldots, H_{\ell-1}/H_\ell)$, these original lists are themselves bijective up to isomorphism. In particular $k = \ell$ and the multisets of composition factors agree. This completes the uniqueness proof, and hence the proof of the Jordan–Hölder theorem. [guided] We now harvest the uniqueness statement from Schreier. Apply Schreier's refinement theorem to two composition series \begin{align*} G &= G_0 \triangleright G_1 \triangleright \cdots \triangleright G_k = \{e\}, \\ G &= H_0 \triangleright H_1 \triangleright \cdots \triangleright H_\ell = \{e\}. \end{align*} We obtain refinements of equal length whose factors match up to isomorphism after some permutation. The remaining task is to argue that "refining a composition series" cannot really insert anything new: any insertion must duplicate an existing term, contributing only a $\{e\}$-factor. Why? Suppose the refinement inserts a subgroup $K$ between $G_{i+1}$ and $G_i$, so $G_{i+1} \trianglelefteq K \trianglelefteq G_i$. The factor $G_i/G_{i+1}$ is simple by the composition-series hypothesis. By the **Lattice Isomorphism Theorem** applied to the quotient $G_i \twoheadrightarrow G_i/G_{i+1}$, normal subgroups of $G_i/G_{i+1}$ correspond bijectively to normal subgroups of $G_i$ containing $G_{i+1}$. The image of $K$ under this correspondence is the normal subgroup $K/G_{i+1}$ of the simple group $G_i/G_{i+1}$. By simplicity, $K/G_{i+1}$ is either $\{e\}$ (so $K = G_{i+1}$) or all of $G_i/G_{i+1}$ (so $K = G_i$). Either way, the "new" subgroup duplicates an existing one, and one of the two new factors $G_i/K$, $K/G_{i+1}$ is trivial. So both refinements are obtained from the original composition series by inserting $\{e\}$-factors. The list of *non-trivial* factors of each refinement coincides with the list of factors of the corresponding original composition series. Schreier matches the refinements' factor lists bijectively up to isomorphism, and $\{e\}$-factors match $\{e\}$-factors, so the non-$\{e\}$-factors also match bijectively up to isomorphism. Therefore the original composition series have the same length $k = \ell$ and equivalent factor lists. This proves the Jordan–Hölder theorem. [/guided] [/step]