Series of Groups

Commutators

In [Algebraic Structures] §Abelian Groups, ⁋Definition 3, for any group \(G\) and two subgroups \(H,H'\) of \(G\), we defined the subgroup \([H,H']\) formed by their commutators. In that post, we only considered the case where \(H=H'=G\), but since we will generalize this in the present post, let us first examine the properties of commutators in more detail.

First, from \([h,h']^{-1}=[h',h]\), it always holds that \([H,H']=[H',H]\). If \([H,H']=\{e\}\), then for all \(h,h'\) we have \(hh'=h'h\), so \(H'\subseteq C_G(H)\) and \(H\subseteq C_G(H')\), and it is also clear that if either of these two conditions holds, then \([H,H']=\{e\}\) holds. Similarly, if \([H,H']\subseteq H\), then for any \(h,h'\),

\[h^{-1}h'^{-1}hh'\in H\implies h'^{-1}hh'\in H\]

so \(H'\subseteq N_G(H)\), and the converse is likewise clear. Finally, if \(H,H'\) were both normal subgroups of \(G\), then for any generator \([h,h']=h^{-1}h'^{-1}hh'\) of \([H,H']\) and any element \(x\) of \(G\),

\[x[h,h']x^{-1}=(xh^{-1}x^{-1})(xh'^{-1}x^{-1})(xhx^{-1})(xh'x^{-1})\]

and here, from the assumption that each of \(H, H'\) is a normal subgroup, \([H,H']\) is also a normal subgroup of \(G\).

We summarize the results for this kind of calculation in Lemma 1. Before that, to simplify notation, for an inner automorphism \(\rho_g: x\mapsto gxg^{-1}\), let us write

\[\rho_{g^{-1}}(x)=x^g\]

Then by [Algebraic Structures] §Group Action, ⁋Proposition 9, for any \(g_1,g_2\in G\) and \(x\in G\),

\[(x^{g_1})^{g_2}=x^{g_1g_2}\]

holds.

Lemma 1 For any \(x,y,z\in G\), the following hold.

1. \(xy=yx[x,y]\).
2. \(x^y=[y,x^{-1}]x\).
3. \([x,yz]=[x,z][x,y]^z=[x,z][z,[y,x]][x,y]\).
4. \([xy,z]=[x,z]^y[y,z]=[x,z][[x,z],y][y,z]\).
5. \([x^y,[y,z]][y^z,[z,x]][z^x,[x,y]]=e\).
6. \([x,yz][z,xy][y,zx]=e\).
7. \([xy,z][yz,x][zx,y]=e\).

The proof of this is simply expanding each side, so we omit it. Using this, we can show the following.

Proposition 2 For a group \(G\) and three subgroups \(H, H', H''\), the following hold.

1. \(H\) normalizes \([H,H']\).
2. If \([H',H'']\) normalizes \(H\), then \([H,[H',H'']]\) is equal to the subgroup generated by elements of the form \([h,[h',h'']]\) in \(G\).

3. If \(H, H', H''\) are all normal, then the inclusion

   \[[H, [H',H'']]\subseteq[H'',[H',H]][H', [H'',H]]\]

   holds.

Proof

1. It suffices to show that for any generator \([h_1,h']\in [H,H']\) and \(h_2\in H\), we have \([h_1,h']^{h_2}\in [H,H']\). This follows from the fourth result of Lemma 1, since

   \[[h_1,h']^{h_2}=[h_1h_2,h'][h_2,h']^{-1}\]

   holds.

2. For convenience, let the subgroup generated by elements of the form \([h,[h',h'']]\) in \(G\) be \(K\). Then since \(K\subseteq [H, [H',H'']]\) is clear, what we need to show is the reverse inclusion \([H,[H',H'']]\subseteq K\). Now since \([H, [H',H'']]\) is generated by elements of the form \([h, [h_1', h_1'']\cdots[h_k',h_k'']]\), what we need to show is that elements of this form belong to \(K\).
   In general, fixing \(h\in H, h'\in H', h''\in H''\) and \(x\in G\), consider the element \([h, [h',h'']x]\). Then from the third result of Lemma 1,

   \[[h, [h',h'']x]=[h,x][h, [h', h'']]^x=[h,x][x,[[h',h''],h]][h, [h',h'']]\]

   holds. Now, from the assumption that \([H',H'']\) normalizes \(H\), since \([[h',h''],h]\) is an element of \(H\), if \(x\) is an element of \([H', H'']\), then each term in the above expression is an element of \(K\). Therefore, by induction we can show that any generator of \([H, [H',H'']]\) belongs to \(K\).

3. Finally, if \(H, H', H''\) are normal, all groups appearing in the third result are also normal subgroups. Therefore, by the second result, it suffices to show that for any \(h,h',h''\), the element \([h,[h',h'']]\) belongs to the right-hand side. Now, if we set \(u=h^{(h')^{-1}}\), from the equation

   \[[h, [h', h'']]=[u^{h'}, [h', h'']]=[(h'')^u, [u, h']]^{-1}[(h')^{h''},[h'',u]]^{-1}\]

   we obtain the desired result.

Lower Central Series and Nilpotent Groups

Definition 3 For a group \(G\), through the equations

\[C_1(G)=G,\qquad C_{n+1}(G)=[G, C_n(G)]\]

we define the lower central series of \(G\), \((C_n(G))_{n\geq 1}\).

Then the following holds.

Proposition 4 For a group homomorphism \(f:G \rightarrow G'\), \(f(C_n(G))\subseteq C_n(G')\) always holds. Moreover, if \(f\) is surjective, then \(f(C_n(G))=C_n(G')\) holds.

Proof

We proceed by induction on \(n\). Suppose that \(f(C_n(G))\subseteq C_n(G')\). Then by definition, the elements of \(C_{n+1}(G)\) are generated by elements of the form

\[x^{-1}y^{-1}xy,\qquad x\in G, y\in C_n(G)\]

and since

\[f(x^{-1}y^{-1}xy)=f(x)^{-1}f(y)^{-1}f(x)f(y)\in [G', f(C_n(G))]\subseteq [G', C_n(G')]=C_{n+1}(G')\]

we obtain the desired result. If \(f\) is surjective, then \(f(G)=G'\), and in the above induction we can replace \(\subseteq\) with \(=\).

Meanwhile, since \(G\) is a normal subgroup of \(G\), by a similar induction, the \(C_n(G)\) are all normal subgroups of \(G\), and therefore \(C_{n+1}(G)\) is also a normal subgroup of \(C_n(G)\).

Proposition 5 For any group \(G\) and any natural numbers \(m,n\), the inclusion

\[[C_m(G), C_n(G)]\subseteq C_{m+n}(G)\]

holds.

Proof

From the third result of Proposition 2, we know that

\[[C_m(G), C_{n+1}(G)]=[C_m(G), [C_n(G), G]]\subseteq[G, [C_m(G), C_n(G)]][C_n(G), [G, C_m(G)]]=[G, [C_m(G), C_n(G)]][C_n(G), C_{m+1}(G)]\]

holds. Therefore, if \([C_m(G), C_n(G)]\subseteq C_{m+n}(G)\) and \([C_n(G), C_{m+1}(G)]\subseteq C_{m+n+1}(G)\) hold, then \([C_m(G), C_{n+1}(G)]\subseteq C_{m+n+1}(G)\) also holds. Now, for any \(m\) and \(n\), since it is clear from the definition that

\[[C_m(G), C_1(G)]\subseteq C_{m+1}(G),\qquad [C_1(G), C_n(G)]\subseteq C_{n+1}(G)\]

hold, by induction we see that this inclusion holds for all \(m,n\).

Now we define the following.

Definition 6 A group \(G\) is called a nilpotent group if there exists a natural number \(n\) such that \(C_{n+1}(G)=\{e\}\). The smallest natural number \(n\) satisfying this condition is called the nilpotency class of \(G\).

Then the following holds.

Proposition 7 For a group \(G\) and a natural number \(n\), the following are all equivalent.

1. \(G\) is a nilpotent group of nilpotency class \(\leq n\).

2. There exists a decreasing sequence of subgroups of \(G\)

   \[G=G_1\supset G_2\supset\cdots\supset G_{n+1}=\{e\}\]

   such that \([G, G_k]\subseteq G_{k+1}\) holds for all \(k\).

3. In \(G\), the center \(Z(G)\) contains a subgroup \(A\) such that \(G/A\) is a nilpotent group of nilpotency class \(\leq n-1\).

Proof

First, assuming the first condition, \(G_k=C_k(G)\) satisfies the second condition; conversely, if the second condition holds, we can show by induction that \(C_k(G)\subseteq G_k\) always holds.

For the remaining equivalence, assuming the first condition, the third condition follows by setting \(A=C_n(G)\). To show that the third condition implies the first condition, it suffices to check that under the canonical morphism \(G \rightarrow G/A\), the image of \(C_n(G)\) is equal to \(C_n(G/A)\) by Proposition 4, and since this is \(\{e\}\) by assumption, \(C_n(G)\subseteq A\) and therefore \(C_{n+1}(G)=\{e\}\).

Therefore, intuitively, a nilpotent group of nilpotency class \(\leq n\) can be thought of as being obtained from the trivial group \(\{e\}\) through \(n\) central extensions.

Proposition 8 Fix a nilpotent group \(G\) of nilpotency class \(\leq n\) and, in \(G\), a subgroup \(H\). Then there exists a sequence of subgroups

\[G=H_1\supseteq H_2\supseteq\cdots\supseteq H_{n+1}=H\]

such that \(H_{k+1}\) is a normal subgroup of \(H_k\) and \(H_k/H_{k+1}\) is commutative.

Proof

After choosing a sequence of subgroups of \(G\) satisfying the second equivalent condition of Proposition 7, it suffices to set \(H_k=HG_k\).

Derived Series and Solvable Groups

Now we define another kind of series.

Definition 9 The derived series of a group \(G\) is given by

\[D_0(G)=G,\qquad D_{n+1}(G)=[D_n(G),D_n(G)]\]

as a series of subgroups of \(G\).

Then, just as for nilpotent groups, the following proposition holds.

Proposition 10 For a group homomorphism \(f:G \rightarrow G'\), \(f(D_n(G))\subseteq D_n(G')\) always holds. Furthermore, if \(f\) is surjective, then \(f(D_n(G))=D_n(G')\) holds.

The proof of this can be done by induction, just as in Proposition 4. Just as nilpotent groups were defined by the stability condition of the lower central series, solvable groups are defined by the stability condition of the derived series.

Definition 11 A group \(G\) is called solvable if there exists a natural number \(n\) such that \(D_n(G)=\{e\}\). The smallest natural number \(n\) satisfying this condition is called the solvability class of \(G\).

By definition, \(D_0(G)=C_1(G)=G\) and \(D_1(G)=[G,G]=C_2(G)\) hold. Then from this fact and Proposition 5, we see by induction that the inclusion

\[D_n(G)\subseteq C_{2^n}(G)\]

holds. That is, any nilpotent group is always solvable. The following proposition is the characterization of solvable groups corresponding to Proposition 7.

Proposition 12 For a group \(G\) and a natural number \(n\), the following are all equivalent.

1. \(G\) is a solvable group of solvability class \(\leq n\).

2. There exists a decreasing sequence of subgroups of \(G\)

   \[G=G_1\supset G_2\supset\cdots\supset G_{n+1}=\{e\}\]

   such that the \(G_k/G_{k+1}\) are all commutative.

3. \(G\) has a normal commutative subgroup \(A\) such that \(G/A\) is a solvable group of solvability class \(\leq n-1\).

Therefore, intuitively, a solvable group of solvability class \(\leq n\) can be thought of as being obtained by extending the trivial group \(\{e\}\) by \(n\) abelian groups.

Proposition 13 Subgroups and quotient groups of a solvable group are solvable.

Proof

Let \(G\) be solvable so that \(D_n(G)=\{e\}\). For a subgroup \(H\subseteq G\), applying Proposition 10 to the inclusion \(H \hookrightarrow G\), we have \(D_n(H)\subseteq D_n(G)=\{e\}\). Meanwhile, for a surjection \(\pi:G \rightarrow Q\), by the same proposition we have \(D_n(Q)=\pi(D_n(G))=\{e\}\). (Definition 11)

Proposition 14 For a finite solvable group \(G\), there exists a sequence of subgroups

\[G=H_0\supseteq H_1\supseteq\cdots\supseteq H_r=\{e\}\]

such that each \(H_{i+1}\) is a normal subgroup of \(H_i\) and \(H_i/H_{i+1}\) is a cyclic group.

Proof

First, we show that a finite abelian group \(A\) has such a series by induction on the order of \(A\). If \(A\) is trivial, this is clear. If \(A\) is nontrivial, choose from \(A\) a non-identity element \(a\in A\). Since the order of the quotient group \(A/\langle a\rangle\) is smaller than that of \(A\), by the induction hypothesis there exists a series

\[A/\langle a\rangle=Q_0\supseteq Q_1\supseteq\cdots\supseteq Q_s=\{e\}\]

such that each \(Q_i/Q_{i+1}\) is cyclic. By [Algebraic Structures] §Group Isomorphisms, ⁋Theorem 7 (The fourth isomorphism theorem), each \(Q_i\) can, for a subgroup containing \(\langle a\rangle\) of \(A\), \(H_i\), be uniquely written in the form \(Q_i=H_i/\langle a\rangle\), and since this correspondence preserves inclusion, \(A=H_0\supseteq H_1\supseteq\cdots\supseteq H_s=\langle a\rangle\). Since \(A\) is abelian, each \(H_{i+1}\) is a normal subgroup of \(H_i\), and by [Algebraic Structures] §Group Isomorphisms, ⁋Theorem 6 (The third isomorphism theorem),

\[H_i/H_{i+1}\cong(H_i/\langle a\rangle)/(H_{i+1}/\langle a\rangle)=Q_i/Q_{i+1}\]

is cyclic. Appending \(\langle a\rangle\supseteq\{e\}\) to the end of this series, the final quotient \(\langle a\rangle\) is also cyclic, and thus \(A\) has the desired series.

Now for a general finite solvable group \(G\), if we consider the derived series

\[G=D_0(G)\supseteq D_1(G)\supseteq\cdots\supseteq D_n(G)=\{e\}\]

each \(D_{k+1}(G)=[D_k(G),D_k(G)]\) is a normal subgroup of \(D_k(G)\), and the quotient \(D_k(G)/D_{k+1}(G)\) is a finite abelian group. Therefore, by what was shown above, \(D_k(G)/D_{k+1}(G)\) has a series whose successive quotients are cyclic, and by the same reasoning as lifting the series of \(A/\langle a\rangle\) to \(A\) in the previous paragraph, we obtain a sequence of subgroups

\[D_k(G)=K_0\supseteq K_1\supseteq\cdots\supseteq K_t=D_{k+1}(G)\]

Here, the fact that each \(K_{j+1}\) is a normal subgroup of \(K_j\) follows from the fact that \(D_k(G)/D_{k+1}(G)\) is abelian and that the bijection in [Algebraic Structures] §Group Isomorphisms, ⁋Theorem 7 (The fourth isomorphism theorem) preserves normality, and \(K_j/K_{j+1}\) is cyclic. Inserting these sequences between each step of the derived series yields the desired series.

Composition Series and the Jordan–Hölder Theorem

Earlier, we saw that nilpotent groups are groups obtained by repeated central extensions, and solvable groups are groups obtained by repeated abelian extensions. We now treat the most general case.

Definition 15 A sequence of subgroups of a group \(G\)

\[G=G_0\supset G_1\supset \cdots\supset G_n=\{e\}\]

is called a subnormal series if for each \(k\), \(G_{k+1}\) is a normal subgroup of \(G_k\), in which case \(G_k/G_{k+1}\) is called a quotient of this series. If there is no subnormal series finer than the subnormal series \(G_\bullet\), it is called a composition series of \(G\).

Then, for any group \(G\) and normal subgroup \(N\), since there exists a bijection between the normal subgroups of \(G/N\) and the normal subgroups of \(G\) containing \(N\), saying that \(G_\bullet\) is a composition series is equivalent to saying that for each \(k\), \(G_k/G_{k+1}\) is simple. (§Symmetric Groups, ⁋Definition 12)

If, for two subnormal series

\[G=G_0\supset G_1\supset \cdots\supset G_n=\{e\},\qquad G=H_0\supset H_1\supset\cdots\supset H_m=\{e\}\]

we have \(m=n\) and \(G_k/G_{k+1}\cong H_{\sigma(k)}/H_{\sigma(k)+1}\) holds for all \(k=0,\ldots, n-1\) for some \(\sigma\in S_n\), then \(G_\bullet\) and \(H_\bullet\) are called equivalent subnormal series. The main theorem of this section is Theorem 18 (Jordan-Hölder), which states that if two composition series of a group \(G\) exist, they are equivalent. To prove this, let us begin with the following lemma.

Lemma 16 (Zassenhaus) Let \(G\) be a group with two subgroups \(H,K\), and let \(H',K'\) be normal subgroups of each, respectively. Then \(H'(H\cap K')\) is a normal subgroup of \(H'(H\cap K)\), \(K'(K\cap H')\) is a normal subgroup of \(K'(K\cap H)\), and the following isomorphism

\[\frac{H'(H\cap K)}{H'(H\cap K')}\cong \frac{K'(K\cap H)}{K'(K\cap H')}\]

exists.

Proof

This proof can be roughly summarized by the following lattice:

First, the fact that \(H'\cap K=H'\cap (H\cap K)\) and \(K'\cap H=K'\cap (K\cap H)\) are each a normal subgroup of \(H\cap K\) follows from [Algebraic Structures] §Group Isomorphisms, ⁋Lemma 4. Thus, we know that their product \((H'\cap K)(K'\cap H)\) is also a normal subgroup of \(H\cap K\). Now, looking at the left-hand side of the claimed isomorphism, from [Algebraic Structures] §Group Isomorphisms, ⁋Theorem 5 (The second isomorphism theorem),

\[H'(H'\cap K)(K'\cap H)=H'(H\cap K')\]

is a normal subgroup of \(H'(H\cap K)\), and we know that an isomorphism

\[\frac{H'(H\cap K)}{H'(H\cap K')}\cong \frac{H\cap K}{(H'\cap K)(K'\cap H)}\]

exists.

Then the following holds.

Proposition 17 (Schreier) For any two subnormal series

\[G=G_0\supset G_1\supset \cdots\supset G_n=\{e\},\qquad G=H_0\supset H_1\supset\cdots\supset H_m=\{e\}\]

there exist refinements \(G_\bullet', H_\bullet'\) of each such that the two are equivalent.

Proof

Between \(G_i\) and \(G_{i+1}\), we insert \(G_{i+1}(G_i\cap H_j)\) by varying \(j\), and between \(H_j\) and \(H_{j+1}\), we insert \(H_{j+1}(H_j\cap G_i)\) by varying \(i\); then one can show using Lemma 16 (Zassenhaus) that the refinements thus constructed are equivalent to each other.

Therefore, the following holds.

Theorem 18 (Jordan-Hölder) Any two composition series are equivalent.

Definition 19 We define the length of a group \(G\) as the supremum of the lengths of strictly descending subnormal series.

Then, if \(G\) has a composition series, we know that the length of a composition series of \(G\) is precisely the length of \(G\). Therefore, for \(G\) and its normal subgroup \(N\), the equality

\[\length(G)=\length(G/N)+\length(N)\]

is a consequence of [Algebraic Structures] §Group Isomorphisms, ⁋Theorem 7 (The fourth isomorphism theorem) and Theorem 18 (Jordan-Hölder).

Solvability of Symmetric Groups

The most classical application of solvability is determining when symmetric groups cease to be solvable, and the turning point lies in the fact that \(A_5\) is simple (§Symmetric Groups, ⁋Example 13).

Proposition 20 \(A_5\) is not a solvable group.

Proof

First, we show that \(A_5\) is non-abelian. If \(A_5\) were abelian, then for any \(g\neq e\), the cyclic subgroup \(\langle g\rangle\) would be a normal subgroup (as is every subgroup of an abelian group) and not \(\{e\}\), so by the simplicity from §Symmetric Groups, ⁋Example 13, we would have \(\langle g\rangle=A_5\); that is, \(A_5\) would be a cyclic group of order \(60\). But then \(\langle g^{30}\rangle\) would be a non-trivial proper normal subgroup of order \(2\), contradicting simplicity. Therefore, \(A_5\) is non-abelian.

Now consider the derived series from Definition 9. The commutator subgroup \(D_1(A_5)=[A_5,A_5]\) is always a normal subgroup, and since \(A_5\) is non-abelian, \(D_1(A_5)\neq\{e\}\). Because \(A_5\) is simple, we must have \(D_1(A_5)=A_5\), and thus inductively, for all \(n\), we have \(D_n(A_5)=A_5\neq\{e\}\). Since there is no \(n\) for which \(D_{n+1}(A_5)=\{e\}\) holds, by Definition 11, \(A_5\) is not a solvable group.

Combined with the fact that solvability is inherited by subgroups, the conclusion regarding the solvability of symmetric groups follows immediately.

Corollary 21 If \(n\geq 5\), then the symmetric group \(S_n\) is not a solvable group.

Proof

If \(n\geq 5\), the permutations fixing the elements of \(\{6,\ldots,n\}\) form a copy of \(S_5\), so \(A_5\subseteq S_5\subseteq S_n\). If \(S_n\) were solvable, then by Proposition 13, its subgroup \(A_5\) would also have to be solvable, which contradicts Proposition 20. Therefore, \(S_n\) is not a solvable group.

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References

[DF] D. S. Dummit and R. M. Foote, Abstract algebra, 3rd ed., Wiley, 2004.