Series of a Group

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Commutators

We defined, in §Abelian Groups, ⁋Definition 3, for any group \(G\) and any two subgroups \(H, H'\) of \(G\), the subgroup \([H, H']\) generated by their commutators. In that post we only considered the case \(H = H' = G\), 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 follows that \([H, H'] = [H', H]\). If \([H, H'] = \{e\}\), then \(hh' = h'h\) for all \(h, h'\), so \(C_G(H) \subseteq H'\) and \(C_G(H') \subseteq H\); conversely, if either of these two conditions holds then \([H, H'] = \{e\}\) is immediate. 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 immediate. Finally, if \(H\) and \(H'\) are 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 from the assumption that \(H\) and \(H'\) are each normal it follows that \([H, H']\) is also a normal subgroup of \(G\).

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

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

Then by §Group Actions, ⁋Proposition 9, for any \(g_1, g_2 \in G\) and \(x \in G\),

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

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 consists merely of expanding each side, so we omit it. Using this we can show the following.

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

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

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

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

   holds.

Proof

1. For any generator \([h_1, h'] \in [H, H']\) and any \(h_2 \in H\) it suffices to show \([h_1, h']^{h_2} \in [H, H']\). This follows from the fourth result of Lemma 1:

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

2. For convenience let \(K\) denote the subgroup of \(G\) generated by elements of the form \([h, [h', h'']]\). Then \(K \subseteq [H, [H', H'']]\) is obvious, so what we must show is the reverse inclusion \([H, [H', H'']] \subseteq K\). Now \([H, [H', H'']]\) is generated by elements of the form \([h, [h_1', h_1''] \cdots [h_k', h_k'']]\), so we must show that elements of this form lie in \(K\). In general, fix \(h \in H\), \(h' \in H'\), \(h'' \in H''\) and \(x \in G\), and consider the element \([h, [h', h'']x]\). Then by the third result of Lemma 1,

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

   From the assumption that \([H', H'']\) normalizes \(H\) it follows that \([[h', h''], h]\) is an element of \(H\); thus if \(x\) were an element of \([H', H'']\) then each term in the above expression would be an element of \([H, [H', H'']]\). Therefore by induction we can show that any generator of \([H, [H', H'']]\) lies in \(K\).

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

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

   we obtain the desired result.

Lower Central Series and Nilpotent Groups

Definition 3 For a group \(G\), we define the lower central series \((C_n(G))_{n \geq 1}\) of \(G\) by the formula

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

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 \(n\). Assume \(f(C_n(G)) \subseteq C_n(G')\). Then 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

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

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

On the other hand, since \(G\) is a normal subgroup of itself, by the same induction we see that 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

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

holds.

Proof

From the third result of Proposition 2 we know

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

Thus 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 \(n\),

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

hold by definition, so by induction we know this inequality holds for all \(m, n\).

We now 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 largest such \(n\) 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. There exists a subgroup \(A\) contained in the center \(C(G)\) of \(G\) 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 then by induction we can show \(C_k(G) \subset G_k\) always holds.

For the remaining equivalence, assuming the first condition, the third condition holds by taking \(A = C_n(G)\). To show the first condition holds assuming the third, we send \(C_n(G)\) via the canonical morphism \(G \rightarrow G/A\); then by Proposition 4 its image equals \(C_n(G/A)\), which is \(\{e\}\) by assumption, so \(C_n(G) \subset A\) and therefore \(C_{n+1}(G) = \{e\}\).

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

Proposition 8 Fix a nilpotent group \(G\) of nilpotency class \(\leq n\) and a subgroup \(H\) of \(G\). 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

Take a sequence of subgroups of \(G\) satisfying the second equivalent condition of Proposition 7, and set \(H_k = HG_k\).

Derived Series and Solvable Groups

We now define another kind of series.

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

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

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. Moreover, if \(f\) is surjective then \(f(D_n(G)) = D_n(G')\) holds.

The proof of this is the same as for Proposition 4, using induction. 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 \(D_{n+1}(G) = \{e\}\). The largest such \(n\) 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. From this fact and Proposition 5, by induction we know the inclusion

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

holds. That is, every 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 all \(G_k/G_{k+1}\) are commutative.

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

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

Composition Series and the Jordan-Hölder Theorem

We have verified above that nilpotent groups are those obtained by repeated central extensions, and solvable groups are those obtained by repeated abelian extensions. We now treat the most general case.

Definition 13 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 this case \(G_k/G_{k+1}\) is called a quotient of this series. If no subnormal series finer than a composition series \(G_\bullet\) exists, we call it a composition series of \(G\).

Then for any group \(G\) and any normal subgroup \(N\), since there is a one-to-one correspondence between normal subgroups of \(G/N\) and normal subgroups of \(G\) containing \(N\), the condition that \(G_\bullet\) is a composition series is equivalent to \(G_k/G_{k+1}\) being simple for each \(k\). (§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 there exists \(\sigma \in S_n\) such that \(G_k/G_{k+1} \cong H_{\sigma(k)}/H_{\sigma(k)+1}\) for all \(k = 0, \ldots, n-1\), then \(G_\bullet\) and \(H_\bullet\) are called equivalent subnormal series. The main theorem of this section is Theorem 16 (Jordan-Hölder), which states that if two composition series of a group \(G\) exist then they are equivalent. To prove this we begin with the following lemma.

Lemma 14 (Zassenhaus) Let \(H, K\) be two subgroups of a group \(G\), and let \(H', K'\) be normal subgroups of \(H\) and \(K\) respectively. Then \(H'(H \cap K')\) is a normal subgroup of \(H'(H \cap K)\), and \(K'(K \cap H')\) 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 summarized roughly by the following lattice:

First, \(H' \cap K = H' \cap (H \cap K)\) and \(K' \cap H = K' \cap (K \cap H)\) are each normal subgroups of \(H \cap K\) by the result of §Isomorphism Theorems, ⁋Lemma 4. Therefore their intersection \((H' \cap K)(K' \cap H)\) is also a normal subgroup of \(H \cap K\). Now looking at the left-hand side of the claimed isomorphism, from §Isomorphism Theorems, ⁋Theorem 5 (The Second Isomorphism Theorem) we know that

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

is a normal subgroup of \(H'(H \cap K)\) and the 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 15 (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 these that are equivalent.

Proof

Insert \(G_i \cap H_j\) between \(G_i\) and \(G_{i+1}\) varying \(j\), and insert \(G_i \cap H_j\) between \(H_j\) and \(H_{j+1}\) varying \(i\); then the refinements so constructed can be shown to be equivalent to each other via Lemma 14 (Zassenhaus).

Therefore the following holds.

Theorem 16 (Jordan-Hölder) Any two composition series are equivalent.

Definition 17 The length of a group \(G\) is defined as the upper bound of the lengths of strictly descending subnormal series.

Then if \(G\) has a composition series, we know that the length of its composition series is exactly the length of \(G\). Therefore the equality

\[\operatorname{length}(G) = \operatorname{length}(G/N) + \operatorname{length}(N)\]

is a consequence of §Isomorphism Theorems, ⁋Theorem 7 (The Fourth Isomorphism Theorem) and Theorem 16 (Jordan-Hölder).

Solvability of Symmetric Groups

The most classical application of solvability is determining from what point symmetric groups cease to be solvable, and the branching point lies in the simplicity of \(A_5\) (§Symmetric Groups, ⁋Example 13).

Proposition 18 \(A_5\) is not solvable.

Proof

First we observe 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 all subgroups of an abelian group are) and not \(\{e\}\), so by the simplicity of \(A_5\) (§Symmetric Groups, ⁋Example 13) we would have \(\langle g \rangle = A_5\), i.e. \(A_5\) would be a cyclic group of order \(60\). But then \(\langle g^{30} \rangle\) would form a nontrivial proper normal subgroup of order \(2\), contradicting simplicity. Therefore \(A_5\) is non-abelian.

Now consider the derived series (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 we have \(D_1(A_5) \neq \{e\}\). Since \(A_5\) is simple we must have \(D_1(A_5) = A_5\), and thus by induction \(D_n(A_5) = A_5 \neq \{e\}\) for all \(n\). Since \(D_{n+1}(A_5) = \{e\}\) cannot hold for any \(n\), by Definition 11 \(A_5\) is not solvable.

Combining this with the fact that solvability is inherited by subgroups, the conclusion about solvability of symmetric groups follows immediately.

Corollary 19 If \(n \geq 5\) then the symmetric group \(S_n\) is not solvable.

Proof

First we observe that a subgroup of a solvable group is again solvable. Applying Proposition 10 to the inclusion \(\iota: H \hookrightarrow G\) of a subgroup \(H \subseteq G\) gives \(D_n(H) \subseteq D_n(G)\), so if \(G\) is solvable and \(D_n(G) = \{e\}\) for some \(n\) then \(D_n(H) = \{e\}\) and \(H\) is also solvable.

Now if \(n \geq 5\), the permutations fixing the elements \(\{6, \ldots, n\}\) form a copy of \(S_5\), so \(A_5 \subseteq S_5 \subseteq S_n\). If \(S_n\) were solvable then its subgroup \(A_5\) would also have to be solvable, which contradicts Proposition 18. Therefore \(S_n\) is not solvable.

Although this corollary is a purely group-theoretic fact, its most famous consequence lies in field theory. The fact that the Galois group of a general polynomial of degree \(n \geq 5\) is \(S_n\) and that this \(S_n\) is not solvable forms the group-theoretic core of the theorem that general equations of degree \(5\) or higher cannot be solved by radicals ([Field Theory] §Solvability by Radicals).