Composition Series, Solvability and Nilpotence: Examples and Counterexamples

Listing the powers of $x$ shows that $⟨x^2⟩$, $⟨x^4⟩$, $⟨x^3⟩$, and $⟨x^6⟩$ have orders $6,3,4,2$, respectively; [L1] confirms that all displayed terms are cyclic subgroups.

Each adjacent quotient therefore has prime order: the first list is $2,2,3$, and the second is $3,2,2$. A group of prime order is simple, so both chains are composition series.

Each quotient is cyclic by [L1], hence the two factor lists are $(C_2,C_2,C_3)$ and $(C_3,C_2,C_2)$. Their products both equal $12$ as [L3] requires, and their agreement up to permutation illustrates [L2].

The normal subgroup $A_3$ has order three and quotient $S_3\mathrm{/}{A}_3$ has order two; both factors are cyclic of prime order and simple, so $S_3▹A_3▹1$ is a composition series by [F1].

By [L1], $S_3^{\mathrm{\prime }}=A_3$; the cyclic group $A_3$ is abelian, so $A_3^{\mathrm{\prime }}=1$. Thus the same chain is the derived series and [F2] gives derived length two.

The center of $S_3$ is trivial: a central element must commute with $(12)$ and $(123)$, but direct multiplication shows that none of the five nonidentity permutations commutes with both.

If a nontrivial group has a central series, take its first nontrivial term $H_j$. Then $H_{j-1}=1$ and centrality gives $H_j\le Z(G)$, so the center is nontrivial. By [F3], a nontrivial nilpotent group has such a series; step 1.3 therefore shows that $S_3$ is not nilpotent.

The displayed terms have orders $24,12,4,2,1$. Each is normal in the preceding term: $A_4$ is the sign kernel, $V_4$ is normal in $A_4$, and $K$ is normal in the abelian group $V_4$.

By [L2], $S_4^{\mathrm{\prime }}=A_4$. The quotient $A_4\mathrm{/}{V}_4$ has order three and is abelian, so [L3] gives $A_4^{\mathrm{\prime }}\le V_4$. Direct calculation gives $[(123),(124)]=(12)(34)$; normality of $A_4^{\mathrm{\prime }}$ from [L4] and conjugation by $A_4$ then put all three nonidentity elements of $V_4$ in $A_4^{\mathrm{\prime }}$. Thus $A_4^{\mathrm{\prime }}=V_4$, while $V_4^{\mathrm{\prime }}=1$ because $V_4$ is abelian. The derived series is therefore the displayed chain of length three by [F2].

The adjacent quotient orders are $2,3,2,2$, so the quotients are simple and the chain is a composition series by [F1].

Thus the composition length is four while the derived length is three; the two series measure different features of $S_4$.

The relation $srs=r^{-1}$ gives $[s,r]=sr{s}^{-1}{r}^{-1}=r^{-2}=r^2$. Every commutator is generated by this one because $D_8$ is generated by $r,s$, so ${\gamma }_2(D_8)=⟨r^2⟩$.

No element outside $⟨r^2⟩$ is central: $r,r^3$ do not commute with $s$, and $s{r}^k$ does not commute with $r$. Hence $Z(D_8)=⟨r^2⟩$.

The element $r^2$ commutes with $r$ and $s$, so $⟨r^2⟩\le Z(D_8)$ and ${\gamma }_3(D_8)=[D_8,⟨r^2⟩]=1$.

The quotient $D_8\mathrm{/}⟨r^2⟩$ is abelian, so its center is the whole quotient and [F2] gives $Z_2(D_8)=D_8$.

Steps 1.1 to 2.2 give both asserted central series, and [L1] gives nilpotency class exactly two.

The element $(0,0,0)$ is an identity and $(a,b,c{)}^{-1}=(-a,-b,-c+ab)$.

Both $(xy)z$ and $x(yz)$ have first two coordinates equal to the coordinate sums and third coordinate $c+c^{\mathrm{\prime }}+c^{\mathrm{\prime }\mathrm{\prime }}+a{b}^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }\mathrm{\prime }}+a^{\mathrm{\prime }}{b}^{\mathrm{\prime }\mathrm{\prime }}$, so the operation is associative. Thus [F1] makes $H$ a group.

Direct use of step 1.1 gives $[(a,b,c),(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})]=(0,0,a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b)\mathrm{.}$ Hence every commutator lies in $C:=\{(0,0,c):c\in \mathbb{Z}\}$.

Every element of $C$ commutes with every element of $H$, and $[(1,0,0),(0,c,0)]=(0,0,c)$; therefore $[H,H]=C\le Z(H)$.

By [F2], ${\gamma }_3(H)=[H,C]=1$, while $(0,0,1)\in {\gamma }_2(H)$ is nonidentity. Thus [L1] gives nilpotency class exactly two.

The chain $C_4>C_2>1$ is a composition series with factors $C_2,C_2$, by [F1] and [L3].

The chain $C_2\times C_2>C_2\times 1>1$ is also a composition series with factors $C_2,C_2$, by [F1] and [L2].

The group $C_4$ has an element of order four, while every nonidentity element of $C_2\times C_2$ has order two by coordinatewise multiplication; hence the groups are not isomorphic.

Thus two nonisomorphic finite groups have the same composition factors, refuting the statement without contradicting [L1], which compares two series of one group rather than different groups.

By [L1], the derived series $S_3>A_3>1$ reaches $1$, so $S_3$ is solvable by [F1].

The transpositions $(12)$ and $(23)$ do not commute, since $(12)(23)=(123)$ while $(23)(12)=(132)$.

Thus the solvable group $S_3$ is nonabelian, refuting the statement.

The group $A_3$ is cyclic of order three, and the sign map has kernel $A_3$ and image $C_2$, so [L2] gives $S_3\mathrm{/}{A}_3\cong C_2$.

The center of $S_3$ is trivial: no nonidentity permutation commutes with both $(12)$ and $(123)$. If a nontrivial group had a central series, its first nontrivial term would lie in its center; hence [F2] shows that $S_3$ is not nilpotent.

Both $A_3$ and $S_3\mathrm{/}{A}_3$ are nontrivial abelian groups and hence nilpotent by [F1].

Therefore $1\to A_3\to S_3\to C_2\to 1$ is an extension of nilpotent groups whose middle group is not nilpotent.

The chain $S_4▹A_4▹V_4▹K▹1$ is subnormal: $A_4⊴S_4$, $V_4⊴A_4$, and $K⊴V_4$ because $V_4$ is abelian.

Conjugation by $(123)$ sends $(12)(34)$ to $(23)(14)$ by [L1], and $(23)(14)\mathrm{\notin }K$; hence $K$ is not normal in $S_4$.

By [F1], the displayed chain is subnormal but not normal, refuting the statement.

The subgroup $H$ is characteristic in $A_4$: it consists of the identity together with all elements of order two, a description preserved by every automorphism.

Since $H$ is abelian, every subgroup of $H$, including $K$, is normal in $H$.

Conjugation by $(123)\in A_4$ sends $(12)(34)$ to $(23)(14)$ by [L2], and this element is not in $K$; hence $K⋬A_4$.

Thus $K⊴H\mathrm{char}G$ but $K⋬G$, refuting the statement and showing why [L1] needs characteristicity at both stages.