Composition Series and the Jordan-Holder Theorem

We use strong induction on $|G|$, the technique of induction on group order. If $|G| = 1$, the series $\{e\} = G_0 = G$ (length $0$, no factors) is a composition series vacuously: there is nothing to refine. Suppose $|G| > 1$ and that every group of order strictly less than $|G|$ has a composition series. The set of proper normal subgroups of $G$ is nonempty (it contains $\{e\}$) and finite, so it has a maximal element $N$ with respect to inclusion — a normal subgroup $N \trianglelefteq G$, $N \ne G$, such that no normal subgroup $M$ of $G$ satisfies $N \subsetneq M \subsetneq G$. By the Isomorphism Theorems (specifically the correspondence theorem for the quotient $G/N$), subgroups of $G/N$ containing the identity correspond bijectively, inclusion-preservingly, and normality-preservingly to subgroups of $G$ containing $N$. Since $N$ is maximal among proper normal subgroups of $G$, there is no normal subgroup of $G/N$ strictly between $\{e\}$ and $G/N$. Hence $G/N$ has no proper nontrivial normal subgroup, i.e. $G/N$ is simple (it is nontrivial since $N \ne G$). Since $N \ne G$ and $G$ is finite, $|N| < |G|$, so by the inductive hypothesis $N$ has a composition series

with all factors $N_{i+1}/N_i$ simple. Appending $G$ to this chain,

gives a normal series of $G$ (the new relation $N \trianglelefteq G$ holds by construction) whose factors are $N_1/N_0, \ldots, N_k/N_{k-1}$ (all simple, by the inductive hypothesis) together with the final factor $G/N$ (simple, as shown above). This is a composition series for $G$. By strong induction, every finite group has a composition series. $\blacksquare$

Set $D = A^*\cap B^*$. First note $A^*\cap B \trianglelefteq A^* \cap B^*$: indeed for $x \in A^*\cap B^*$ and $n \in A^*\cap B$, since $B\trianglelefteq B^*$ we get $xnx^{-1}\in B$, and since $x,n\in A^*$ we get $xnx^{-1}\in A^*$, so $xnx^{-1}\in A^*\cap B = A^*\cap B$. Thus $A^*\cap B\trianglelefteq D$. Symmetrically $B^*\cap A \trianglelefteq D$.

Step 1 (products with a normal subgroup are subgroups). Since $A\trianglelefteq A^*$, for any subgroup $H\le A^*$ the set $AH$ is a subgroup of $A^*$ (this is the standard "product of a subgroup with a normal subgroup is a subgroup" fact, using that $aH a^{-1}=H$ implies $AH=HA$). Taking $H = A^*\cap B$ (a subgroup of $A^*$) gives that $A(A^*\cap B)$ is a subgroup of $A^*$; taking $H=D=A^*\cap B^*$ gives that $A(A^*\cap B^*)$ is a subgroup of $A^*$ as well.

Step 2 ($A(A^*\cap B)$ is normal in $A(A^*\cap D)=A\cdot D$). We show $A(A^*\cap B) \trianglelefteq A\cdot D$. Note $A\cdot D = A(A^*\cap B^*)$ since $A\le A^*$. Let $g = a d \in A\cdot D$ with $a\in A$, $d\in D=A^*\cap B^*$, and let $x = a' n$ with $a'\in A$, $n\in A^*\cap B$, be a general element of $A(A^*\cap B)$. Then

Since $A \trianglelefteq A^*$ and $d, a' \in A^*$, we have $d a' d^{-1} \in A$. Since $A^*\cap B \trianglelefteq D$ (shown above) and $d\in D$, $n \in A^*\cap B$, we get $dnd^{-1}\in A^*\cap B$. Hence

because $a(da'd^{-1}) \in A$ and $(dnd^{-1})a^{-1}$, after moving the $A$-part using normality of $A$ in $A^*$ ($a^{-1}$ conjugates the $A^*\cap B$-element... more directly: write the whole expression as a product of an element of $A$ times an element of $A^*\cap B$, which suffices since $A(A^*\cap B)$ consists of exactly such products, and $A$ is normal so we may freely reorder $A$-parts to the front). This proves claim (1): $A(A^*\cap B) \trianglelefteq A(A^*\cap B^*)$. Claim (2) is identical with the roles of $(A,A^*)$ and $(B,B^*)$ exchanged.

Step 3 (the isomorphism). Consider the subgroup $D = A^*\cap B^*$ and the map

This is well-defined since $D = A^*\cap B^* \le A(A^*\cap B^*)$. It is a homomorphism as the restriction of the quotient map. It is surjective: every coset of $A(A^*\cap B)$ inside $A(A^*\cap B^*)$ has a representative of the form $a\cdot d$ with $a \in A \le A(A^*\cap B)$ and $d\in D$, hence equals $\varphi(d)$. Compute the kernel:

We claim $D \cap A(A^*\cap B) = (A\cap B^*)(A^*\cap B) = (A\cap D)(A^*\cap B)$. Indeed if $d = a n \in D$ with $a \in A$, $n \in A^*\cap B\le D$, then $a = dn^{-1}\in D$ (as $d,n \in D$), so $a \in A\cap D$; conversely if $a\in A\cap D$ and $n \in A^*\cap B$ then $an \in A(A^*\cap B)$ and $an \in D$ since $D$ is a group containing both $a,n$. This proves the claimed equality, and one checks similarly that $A\cap D = A\cap B^*$ (both intersections of $A$ with $B^*$, since $A\le A^*$ forces $A \cap A^*\cap B^* = A\cap B^*$). Hence by the First Isomorphism Theorem applied to $\varphi$,

By the identical argument with $A$ and $B$ exchanged,

But $(A\cap B^*)(A^*\cap B) = (B\cap A^*)(B^*\cap A)$ as sets (both equal the subgroup generated by $A\cap B^*$ and $A^*\cap B$, which are literally the same two subgroups written in the other order), so the two left-hand sides are the identical quotient group $D/(A^*\cap B)(A\cap B^*)$. Composing the two isomorphisms (through this common quotient) gives

as claimed. $\blacksquare$

Define, for $0\le i \le m-1$ and $0\le j\le n$,

and for $0\le j\le n-1$ and $0\le i \le m$,

Apply the Zassenhaus lemma to $A = A_i \trianglelefteq A^* = A_{i+1}$ and $B = B_j \trianglelefteq B^* = B_{j+1}$: we obtain

and symmetrically $B_{j,i}\trianglelefteq B_{j,i+1}$, together with the isomorphism

So for each fixed $i$, inserting the terms $A_{i,0}\trianglelefteq A_{i,1}\trianglelefteq\cdots\trianglelefteq A_{i,n}$ refines the single step $A_i\trianglelefteq A_{i+1}$ (note $A_{i,0}=A_i(A_{i+1}\cap B_0)=A_i(A_{i+1}\cap\{e\})=A_i$ and $A_{i,n}=A_i(A_{i+1}\cap B_n)=A_i(A_{i+1}\cap G)=A_{i+1}$). Stacking these refinements for $i=0,\ldots,m-1$ produces a refinement of the entire $A$-series into $mn$ steps, with factors $A_{i,j+1}/A_{i,j}$ for $0\le i< m$, $0\le j<n$. Symmetrically, inserting $B_{j,0}\trianglelefteq\cdots\trianglelefteq B_{j,m}$ into each step $B_j\trianglelefteq B_{j+1}$ produces a refinement of the entire $B$-series into $mn$ steps, with factors $B_{j,i+1}/B_{j,i}$ for $0\le j<n$, $0\le i<m$. Both refinements have exactly $mn$ factors, and the Zassenhaus isomorphism gives a bijection between them — the factor indexed by $(i,j)$ on the $A$-side corresponds to the factor indexed by $(j,i)$ on the $B$-side — under which corresponding factors are isomorphic:

Hence the two refinements are equivalent. $\blacksquare$

By the Schreier Refinement Theorem, the two given series admit refinements $\widetilde{G}_\bullet$ and $\widetilde{H}_\bullet$ that are equivalent — the same multiset of factors up to isomorphism and reordering. We claim a composition series admits no proper refinement other than by repeating terms (inserting trivial factors). Indeed, suppose $\{e\}=G_0\trianglelefteq\cdots\trianglelefteq G_m=G$ is a composition series and a refinement inserts a new term $N$ strictly between some $G_{i-1}$ and $G_i$, so $G_{i-1}\trianglelefteq N \trianglelefteq G_i$ with $N \ne G_{i-1}, G_i$. Since $G_{i-1}\trianglelefteq G_i$, and $N/G_{i-1}$ is by hypothesis a subgroup of $G_i/G_{i-1}$, normal in $G_i/G_{i-1}$ (as $N\trianglelefteq G_i$ implies $N/G_{i-1}\trianglelefteq G_i/G_{i-1}$ by the correspondence theorem), this exhibits a proper nontrivial normal subgroup $N/G_{i-1}$ of $G_i/G_{i-1}$, contradicting that $G_i/G_{i-1}$ is simple. So no such $N$ can exist: every refinement of a composition series can only repeat existing terms, contributing extra trivial factors $\{e\}$, which are conventionally discarded (a "factor" $\{e\}/\{e\}\cong\{e\}$ is not simple and is not counted). Therefore the refinement $\widetilde{G}_\bullet$ of the composition series $G_\bullet$ has, after discarding repeated terms, exactly the same nontrivial factors as $G_\bullet$ itself — namely $G_1/G_0,\ldots,G_m/G_{m-1}$, in the same multiset. Likewise $\widetilde H_\bullet$ has, after discarding repeats, exactly the factors of $H_\bullet$. But by Schreier, $\widetilde G_\bullet$ and $\widetilde H_\bullet$ are equivalent: the same multiset of factors up to isomorphism. Discarding trivial factors from an equivalence of series preserves the equivalence (trivial factors correspond only to trivial factors, since a trivial factor $\{e\}$ is not isomorphic to any simple group). Hence the nontrivial factors of $\widetilde G_\bullet$ — which are exactly the factors of $G_\bullet$ — form the same multiset, up to isomorphism, as the nontrivial factors of $\widetilde H_\bullet$ — which are exactly the factors of $H_\bullet$. This is precisely the statement that $m=n$ and that there is a bijection (permutation) matching the factors of the two composition series up to isomorphism. $\blacksquare$