Composition Series, the Jordan–Hölder Theorem and Solvable Groups

If $\mathrm{\mid }G\mathrm{\mid }=1$, the one-term chain $G=1$ is a composition series of length zero.

Assume $\mathrm{\mid }G\mathrm{\mid }>1$ and that every group of order smaller than $\mathrm{\mid }G\mathrm{\mid }$ has a composition series.

The finite nonempty set of proper normal subgroups of $G$ contains $1$, so choose one, say $N$, of maximum cardinality. This is finite maximization and uses no choice principle.

The quotient $G\mathrm{/}N$ is simple: a nontrivial proper normal subgroup of $G\mathrm{/}N$ would correspond by [L1] to a proper normal subgroup of $G$ strictly containing $N$, contrary to maximality.

Since $N$ is proper, $\mathrm{\mid }N\mathrm{\mid }<\mathrm{\mid }G\mathrm{\mid }$, so the induction hypothesis supplies a composition series $N=N_0▹\cdots ▹N_r=1$.

Prepending $G▹N$ to the series of step 3.2 gives a strict subnormal series whose new factor $G\mathrm{/}N$ is simple by step 3.1 and whose remaining factors are simple by induction; hence it is a composition series of $G$.

If $x\in A(B\cap C)$, write $x=ab$ with $a\in A$ and $b\in B\cap C$; then $x\in AB$, and $a,b\in C$ gives $x\in C$, so $x\in AB\cap C$.

If $x\in AB\cap C$, write $x=ab$ with $a\in A$ and $b\in B$; since $a,x\in C$, one has $b=a^{-1}x\in C$, hence $b\in B\cap C$ and $x\in A(B\cap C)$.

The two inclusions prove $A(B\cap C)=AB\cap C$.

Put $M=A^{\ast }\cap B^{\ast }$, $U=A\cap B^{\ast }$, and $V=A^{\ast }\cap B$. Conjugation by elements of $M$ preserves $U$ and $V$, because it preserves $A,A^{\ast },B,B^{\ast }$; hence $U,V⊴M$ and $D:=UV⊴M$.

The subgroups $X=AV$ and $X^{\ast }=AM$ are well defined. The subgroup $M$ normalizes both $A$ and $V$. Also, for $a\in A$ and $v\in V$, one has $av{a}^{-1}=(av{a}^{-1}{v}^{-1})v\in AV$ because $A⊴A^{\ast }$; hence $A$ normalizes $AV$ and $X⊴X^{\ast }$. The symmetric argument gives $Y=UB⊴MB=Y^{\ast }$.

Apply [L2] inside $X^{\ast }=AM$ with normal subgroup $X=AV$: since $X^{\ast }=XM$, one has $X^{\ast }\mathrm{/}X\cong M\mathrm{/}(M\cap X)$.

Symmetrically, [L2] inside $Y^{\ast }=MB$ with normal subgroup $Y=UB$ gives $Y^{\ast }\mathrm{/}Y\cong M\mathrm{/}(M\cap Y)$, and [L1] gives $M\cap UB=U(M\cap B)=UV=D$.

By [L1], $M\cap AV=(M\cap A)V=UV=D$, so $X^{\ast }\mathrm{/}X\cong M\mathrm{/}D$.

Both quotients are isomorphic to $M\mathrm{/}D$, so $X^{\ast }\mathrm{/}X\cong Y^{\ast }\mathrm{/}Y$; step 2.1 supplies the two normality assertions.

For $0\le i<m$ and $0\le j\le n$, set $G_{i,j}:=G_{i+1}(G_i\cap H_j)$. Then $G_{i,0}=G_i$ and $G_{i,n}=G_{i+1}$; [L1] applied to $G_{i+1}⊴G_i$ and $H_{j+1}⊴H_j$ gives $G_{i,j+1}⊴G_{i,j}$.

For $0\le j<n$ and $0\le i\le m$, set $H_{j,i}:=(G_i\cap H_j)H_{j+1}$. Concatenating these chains gives a subnormal refinement of the $H$-series.

Concatenating the finite chains $G_{i,0}⊵\cdots ⊵G_{i,n}$ over $i=0,\dots ,m-1$ yields a subnormal refinement of the $G$-series, possibly with repeated adjacent terms.

For every cell $(i,j)$, [L1] identifies $G_{i,j}\mathrm{/}{G}_{i,j+1}$ with $H_{j,i}\mathrm{/}{H}_{j,i+1}$. Thus the two refinements have their $mn$ displayed factors paired by $(i,j)\leftrightarrow (j,i)$.

Deleting repeated adjacent terms deletes exactly the trivial factors on both sides of each paired cell, so the remaining factors are still paired and isomorphic; the refinements are equivalent.

By [L1], the two composition series have equivalent refinements.

A subnormal refinement cannot insert a term strictly between adjacent terms $H▹N$: the first inserted term $K$ would satisfy $N<K<H$ and $K⊴H$, so [L2] would make $K\mathrm{/}N$ a nontrivial proper normal subgroup of the simple factor $H\mathrm{/}N$.

Therefore each refinement differs from its original composition series only by repeated adjacent terms, and deleting those repetitions recovers the original series.

Equivalence of the refinements now pairs the original nontrivial factors; hence the two series have the same number of factors, and a permutation matches their factors up to isomorphism.

For every $0\le i<n$, [L1] and [L2] give $\mathrm{\mid }{G}_i\mathrm{\mid }=\mathrm{\mid }{G}_i\mathrm{/}{G}_{i+1}\mathrm{\mid }\mathrm{\mid }{G}_{i+1}\mathrm{\mid }$.

Multiplying the identities of step 1.1 and cancelling the intermediate positive integers gives $\mathrm{\mid }{G}_0\mathrm{\mid }=\left({\prod }_{i=0}^{n-1}\mathrm{\mid }{G}_i\mathrm{/}{G}_{i+1}\mathrm{\mid }\right)\mathrm{\mid }{G}_n\mathrm{\mid }$.

Since $G_0=G$ and $G_n=1$, one has $\mathrm{\mid }{G}_n\mathrm{\mid }=1$, proving the formula. When $n=0$, step 2.1 reads $\mathrm{\mid }1\mathrm{\mid }=1$ by the empty-product convention.

For each $g\in G$, conjugation $x\mapsto gx{g}^{-1}$ is an automorphism of $G$; if $H\mathrm{char}G$, [F1] says it preserves $H$, so [F2] gives $H⊴G$.

Suppose $K\mathrm{char}H\mathrm{char}G$ and let $\alpha \in \mathrm{Aut}(G)$. By [F1], $\alpha (H)=H$, so $\alpha {\mathrm{\mid }}_H$ is an automorphism of $H$; applying [F1] to $K\mathrm{char}H$ gives $\alpha (K)=K$.

Since step 1.2 holds for every automorphism of $G$, $K\mathrm{char}G$; together with step 1.1 this proves both assertions.

Every automorphism $\alpha$ satisfies $\alpha ([x,y])=[\alpha (x),\alpha (y)]$, so it maps the generating commutators of $G^{\mathrm{\prime }}$ into $G^{\mathrm{\prime }}$; applying the same argument to ${\alpha }^{-1}$ gives $\alpha (G^{\mathrm{\prime }})=G^{\mathrm{\prime }}$.

For $x,y\in G$, the group $A$ is abelian, so $f([x,y])=[f(x),f(y)]=1$; hence every generator of $G^{\mathrm{\prime }}$ lies in $\ker f$, and $G^{\mathrm{\prime }}\le \ker f$.

Thus $G^{\mathrm{\prime }}$ is characteristic by [F2], and therefore normal by [L1].

Since $G^{\mathrm{\prime }}\le G^{\mathrm{\prime }}$, [L2] gives that $G\mathrm{/}{G}^{\mathrm{\prime }}$ is abelian.

By [L3] there is a unique $\bar{f}:G\mathrm{/}{G}^{\mathrm{\prime }}\to A$ with $f=\bar{f}\circ q$. Together with step 3.1, this is the asserted universal abelian quotient.

For all $x,y\in G$, $f([x,y])=[f(x),f(y)]$ by expanding the commutator and using [F2].

At $r=0$, $f(G^{(0)})=f(G)=f(G{)}^{(0)}$.

Assume $f(G^{(r)})\le f(G{)}^{(r)}$ and, when $f$ is surjective, $f(G^{(r)})=H^{(r)}$.

Step 1.1 sends every generator of $G^{(r+1)}=[G^{(r)},G^{(r)}]$ into $[f(G{)}^{(r)},f(G{)}^{(r)}]=f(G{)}^{(r+1)}$, proving the inclusion at $r+1$.

If $f$ is surjective and equality holds at $r$, every commutator generator of $H^{(r+1)}$ is the image under $f$ of a commutator of preimages in $G^{(r)}$; thus equality also holds at $r+1$.

Induction gives the inclusion for every $r$, and gives equality for surjective $f$. Applying the inclusion to the inclusion homomorphism $K\hookrightarrow G$ yields $K^{(r)}\le G^{(r)}$.

Suppose $G$ is solvable, and choose $n$ with $G^{(n)}=1$. The derived chain $G=G^{(0)}⊵\cdots ⊵G^{(n)}=1$ is subnormal, and [L2] makes every factor $G^{(i)}\mathrm{/}{G}^{(i+1)}$ abelian.

Conversely, suppose $G=G_0⊵\cdots ⊵G_n=1$ is subnormal with abelian factors. By [L2], $G_i^{\mathrm{\prime }}\le G_{i+1}$ for every $i<n$.

Starting with $G^{(0)}=G_0$, if $G^{(i)}\le G_i$, then [L1] gives $G^{(i+1)}=(G^{(i)}{)}^{\mathrm{\prime }}\le G_i^{\mathrm{\prime }}\le G_{i+1}$; hence $G^{(i)}\le G_i$ for every $i\le n$.

Step 2.1 gives $G^{(n)}\le G_n=1$, so $G$ is solvable by [F2]. Steps 1.1 and 3.1 prove both directions.

Choose $n$ with $G^{(n)}=1$.

By [L1], $H^{(n)}\le G^{(n)}=1$, so $H$ is solvable.

Since the quotient map is surjective, [L1] and [F2] give $(G\mathrm{/}N{)}^{(n)}=q(G^{(n)})=1$, so $G\mathrm{/}N$ is solvable.

Thus solvability passes to both subgroups and quotients.

Choose $r,s$ with $(G\mathrm{/}N{)}^{(r)}=1$ and $N^{(s)}=1$.

Coordinatewise calculation using [L2] gives $[(g,h),(g^{\mathrm{\prime }},h^{\mathrm{\prime }})]=([g,g^{\mathrm{\prime }}],[h,h^{\mathrm{\prime }}])$, so $(G_1\times G_2{)}^{\mathrm{\prime }}=G_1^{\mathrm{\prime }}\times G_2^{\mathrm{\prime }}$ and therefore $(G_1\times G_2{)}^{(k)}=G_1^{(k)}\times G_2^{(k)}$ for every $k$.

By [L1], $q(G^{(r)})=1$, so $G^{(r)}\le N$. Repeatedly applying the subgroup inclusion in [L1] gives $G^{(r+s)}=(G^{(r)}{)}^{(s)}\le N^{(s)}=1$.

Hence $G$ is solvable by [F1].

Choosing a common bound for the derived lengths in a nonempty finite family and using step 1.2 inductively proves its product solvable; for the empty family the product is the trivial group, which has derived length zero.

Suppose $G$ is solvable. Each composition factor is a quotient of a subgroup of $G$, hence is solvable by [L3].

Conversely, take a composition series $G=G_0▹\cdots ▹G_n=1$ whose factors have prime order. The trivial group $G_n$ is solvable, and if $G_{i+1}$ is solvable then $G_i\mathrm{/}{G}_{i+1}$ is cyclic, hence abelian and solvable, so [L4] makes $G_i$ solvable. Finite upward induction gives $G=G_0$ solvable.

Let $S$ be a simple solvable composition factor. Its derived subgroup $S^{\mathrm{\prime }}$ is normal by [L7], so simplicity gives $S^{\mathrm{\prime }}=1$ or $S^{\mathrm{\prime }}=S$; solvability excludes $S^{\mathrm{\prime }}=S$, and therefore $S$ is abelian.

Choose $1\mathrm{\ne }x\in S$. Since $S$ is abelian, $⟨x⟩⊴S$, so simplicity gives $S=⟨x⟩$. By [L6] choose a prime $p$ dividing $\mathrm{\mid }S\mathrm{\mid }$; [L5] gives a subgroup of order $p$, which is nontrivial and normal in the abelian group $S$, hence equals $S$. Thus $S$ is cyclic of prime order.

Steps 1.1, 2.1, and 3.1 prove that solvability forces prime-order composition factors, while step 1.2 proves the converse; [L2] makes the condition independent of the chosen composition series.

By [L3], every positive term of the derived series of $A_n$ equals $A_n$, which is nontrivial by [L2]; hence the series never reaches $1$, and $A_n$ is not solvable by [F1].

Since $A_n\le S_n$, solvability of $S_n$ would imply solvability of $A_n$ by [L1], contradicting step 1.1.

Therefore $A_n$ and $S_n$ are nonsolvable for all $n\ge 5$, including $n=5$.

Suppose $H\mathrm{/}N\le Z(G\mathrm{/}N)$. For every $g\in G$ and $h\in H$, the cosets $gN$ and $hN$ commute by [F2]. Expanding their products with [F3] gives $ghN=hgN$, equivalently $gh{g}^{-1}{h}^{-1}\in N$; hence [F1] gives $[G,H]\le N$.

Conversely, suppose $[G,H]\le N$. Then [F1] gives $gh{g}^{-1}{h}^{-1}\in N$ for all $g\in G,h\in H$. Reversing the coset calculation in step 1.1 shows $gN$ and $hN$ commute, so $H\mathrm{/}N\le Z(G\mathrm{/}N)$ by [F2].

Applying the equivalence of steps 1.1 and 2.1 with $(N,H)=(H_i,H_{i+1})$ for each adjacent pair proves the series criterion, including $H_0=1$ and $H_c=G$.

Let $1=H_0\le \cdots \le H_c=G$ be central. Inductively, $H_i\le Z_i(G)$: the base is $H_0=Z_0(G)=1$, and [L1] says $[G,H_{i+1}]\le H_i\le Z_i(G)$, so the quotient criterion places $H_{i+1}\mathrm{/}{Z}_i(G)$ in $Z(G\mathrm{/}{Z}_i(G))$, hence $H_{i+1}\le Z_{i+1}(G)$.

For any central series as above, descending induction gives ${\gamma }_{c-i+1}(G)\le H_i$: at $i=c$, ${\gamma }_1(G)=G=H_c$; if ${\gamma }_{c-i+1}(G)\le H_i$, then ${\gamma }_{c-i+2}(G)=[G,{\gamma }_{c-i+1}(G)]\le [G,H_i]\le H_{i-1}$ by [L1].

Conversely, if ${\gamma }_{c+1}(G)=1$, the reversed lower central chain $1={\gamma }_{c+1}(G)\le {\gamma }_c(G)\le \cdots \le {\gamma }_1(G)=G$ is central because $[G,{\gamma }_r(G)]={\gamma }_{r+1}(G)$; [L1] applies at every adjacent pair.