Solvable and Nilpotent Groups

(1) Suppose $G/H$ abelian. For $x,y \in G$, in $G/H$ we have $xH \cdot yH = yH \cdot xH$, i.e. $xyH = yxH$, i.e. $(xy)^{-1}(yx) = y^{-1}x^{-1}yx \in H$; taking inverses (a subgroup is closed under inverses) shows $x^{-1}y^{-1}xy = [x,y] \in H$. Since this holds for all generators of $G'$ and $H$ is a subgroup, $G' \le H$. Conversely, if $G' \le H$, then for any $x,y \in G$, $[x,y] \in H$, so $xyH = yxH$ in $G/H$, i.e. cosets $xH$ and $yH$ commute. Since $x,y$ were arbitrary, $G/H$ is abelian.

(2) For any automorphism $\varphi$ of $G$ and generators $[x,y]$ of $G'$, $\varphi([x,y]) = [\varphi(x), \varphi(y)] \in G'$; since $\varphi$ is a bijection and $G'$ is generated by such commutators, $\varphi(G') = G'$. Hence $G'$ is characteristic, and every characteristic subgroup is normal (an inner automorphism is in particular an automorphism). Taking $H = G'$ in part (1) with $G' \le G'$ trivially true, $G/G'$ is abelian.

(3) Since $\varphi$ is a homomorphism, $\varphi([x,y]) = [\varphi(x),\varphi(y)]$ for all $x,y \in G$, so $\varphi(G') \le K'$ (the generators of $G'$ map into generators of $K'$, and homomorphic images of generating sets generate the image). Conversely, since $\varphi$ is surjective, every generator $[\varphi(x),\varphi(y)]$ of $K'$ (as $x,y$ range over $G$, using surjectivity to write any element of $K$ as $\varphi(x)$) is the image $\varphi([x,y])$ of a generator of $G'$, so $K' \le \varphi(G')$. Hence $\varphi(G') = K'$. $\blacksquare$

(a) $\Rightarrow$ (b). Take $G_i = G^{(i)}$. By Basic commutator facts part (2) applied to $G_i$, $G_{i+1} = G_i'$ is normal in $G_i$ and $G_i/G_{i+1}$ is abelian. The chain terminates at $\{e\}$ by hypothesis.

(b) $\Rightarrow$ (a). We show by induction on $i$ that $G^{(i)} \le G_i$. The base case $i=0$ is $G^{(0)} = G = G_0$. Suppose $G^{(i)} \le G_i$. Since $G_i/G_{i+1}$ is abelian, part (1) of Basic commutator facts gives $G_i' \le G_{i+1}$. Since $G^{(i)} \le G_i$, taking commutator subgroups (which is monotone: if $A \le B$ then $A' \le B'$, because commutators of elements of $A$ are commutators of elements of $B$) gives $G^{(i+1)} = (G^{(i)})' \le G_i' \le G_{i+1}$. By induction, $G^{(n)} \le G_n = \{e\}$, so $G^{(n)} = \{e\}$, proving $G$ is solvable.

(i) Subgroups. Let $H \le G$. By monotonicity of the commutator subgroup, if $H \le G$ then $H' \le G'$, and inductively $H^{(i)} \le G^{(i)}$ for all $i$ (same induction as above, using $H^{(i)} \le G^{(i)} \Rightarrow H^{(i+1)} = (H^{(i)})' \le (G^{(i)})' = G^{(i+1)}$). Since $G^{(n)} = \{e\}$ for some $n$, $H^{(n)} \le \{e\}$, so $H^{(n)} = \{e\}$ and $H$ is solvable.

(ii) Quotients. Let $\pi: G \to G/N$ be the quotient map, which is surjective. By Basic commutator facts part (3), $\pi(G^{(i)}) = (G/N)^{(i)}$ for every $i$ — this follows by induction: $\pi(G^{(0)}) = \pi(G) = G/N = (G/N)^{(0)}$, and if $\pi(G^{(i)}) = (G/N)^{(i)}$, then applying part (3) to the surjective homomorphism $\pi|_{G^{(i)}}: G^{(i)} \to (G/N)^{(i)}$ (surjective since $\pi(G^{(i)}) = (G/N)^{(i)}$) gives $\pi(G^{(i+1)}) = \pi((G^{(i)})') = ((G/N)^{(i)})' = (G/N)^{(i+1)}$. If $G^{(n)} = \{e\}$, then $(G/N)^{(n)} = \pi(G^{(n)}) = \pi(\{e\}) = \{e\}$, so $G/N$ is solvable.

(iii) Extensions. Suppose $N$ is solvable, say $N^{(m)} = \{e\}$, and $G/N$ is solvable, say $(G/N)^{(k)} = \{e\}$. By the argument in (ii), $\pi(G^{(k)}) = (G/N)^{(k)} = \{e\}$, so $G^{(k)} \le \ker \pi = N$. Since the derived series is monotone under inclusion (shown in part (i)'s proof: $A \le B \Rightarrow A^{(i)} \le B^{(i)}$), we get $G^{(k+i)} = (G^{(k)})^{(i)} \le N^{(i)}$ for all $i \ge 0$ (applying the derived series of $G^{(k)}$, which by definition of the derived series applied iteratively from $G^{(k)}$ coincides with $(G^{(k)})^{(i)} = G^{(k+i)}$). Taking $i = m$: $G^{(k+m)} \le N^{(m)} = \{e\}$, so $G^{(k+m)} = \{e\}$ and $G$ is solvable. $\blacksquare$

We use throughout the standard facts (from p-Groups and Their Properties) that a nontrivial finite $p$-group has nontrivial center, and (from Composition Series and the Jordan-Holder Theorem background) the Sylow theorems.

(a) Lower and upper central series match up. Claim: for all $i \ge 0$, $\gamma_{i+1}(G) = \{e\}$ implies $Z_i(G) = G$, and conversely. First, note $\gamma_1(G) = G \supseteq \gamma_2(G) \supseteq \cdots$ is a descending chain, each $\gamma_{i+1}(G) = [G,\gamma_i(G)] \trianglelefteq G$ (since for $g \in G$ and a generator $[x,y]$ with $x \in G, y \in \gamma_i(G)$, one computes $g[x,y]g^{-1} = [gxg^{-1}, gyg^{-1}]$ using that $\gamma_i(G)$ is normal — proved by an identical induction to Basic commutator facts(2) — so conjugation permutes the generating commutators of $\gamma_{i+1}(G)$). We show by induction on $i$: $\gamma_{i+1}(G) \le Z_j(G)$ whenever $i + j \ge c$ is not quite the cleanest route; instead we directly show the equivalence by simultaneous induction. Direction 1: Suppose $\gamma_{c+1}(G) = \{e\}$. We show $G/\gamma_{i+1}(G)$ has the property that $\gamma_1(G)/\gamma_{i+1}(G), \ldots$ relates to upper series terms; concretely we show by downward induction on $i$ (from $i=c$ down to $i=0$) that $\gamma_{c-i+1}(G) \le Z_i(G)$. Base case $i = 0$: $\gamma_{c+1}(G) = \{e\} = Z_0(G)$. ✓. Inductive step: suppose $\gamma_{c-i+1}(G) \le Z_i(G)$. We must show $\gamma_{c-i}(G) \le Z_{i+1}(G)$. By definition $Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$, so it suffices to show that for $x \in G$ and $y \in \gamma_{c-i}(G)$, the commutator $[x,y]$ lies in $Z_i(G)$ — because that says $xZ_i(G)$ and $yZ_i(G)$ commute in $G/Z_i(G)$ for every $x \in G$, i.e. $yZ_i(G) \in Z(G/Z_i(G))$, i.e. $y \in Z_{i+1}(G)$. But such commutators $[x,y]$ generate exactly $\gamma_{c-i+1}(G) = [G, \gamma_{c-i}(G)]$, which lies in $Z_i(G)$ by the inductive hypothesis. This proves the step. At $i = c$: $\gamma_1(G) \le Z_c(G)$, i.e. $G \le Z_c(G)$, so $Z_c(G) = G$. Direction 2: Suppose $Z_c(G) = G$. By induction on $i$ from $0$ to $c$ we show $\gamma_{i+1}(G) \le Z_{c-i}(G)$. Base case $i=0$: $\gamma_1(G) = G = Z_c(G)$. ✓. Inductive step: suppose $\gamma_{i+1}(G) \le Z_{c-i}(G)$; since $Z_{c-i}(G)/Z_{c-i-1}(G) = Z(G/Z_{c-i-1}(G))$, for any $x\in G, y \in \gamma_{i+1}(G) \le Z_{c-i}(G)$ we get $[x,y] \in Z_{c-i-1}(G)$ (the commutator of an element with something central mod $Z_{c-i-1}(G)$ lands back in $Z_{c-i-1}(G)$, by definition of centrality of the coset). Hence $\gamma_{i+2}(G) = [G,\gamma_{i+1}(G)] \le Z_{c-i-1}(G)$, completing the induction. At $i = c$: $\gamma_{c+1}(G) \le Z_{-1}(G)$... to avoid a negative index, stop at $i = c-1$: $\gamma_{c}(G) \le Z_1(G)$, and then applying the same one-step argument once more with $x \in G, y \in \gamma_c(G) \le Z_1(G) = Z(G)$ gives $[x,y] = e$ (since $y$ is central), so $\gamma_{c+1}(G) = \{e\}$. Combining both directions proves (a).

(b) Nilpotent $\Rightarrow$ solvable. We show $G^{(i)} \le \gamma_{i+1}(G)$ for all $i \ge 0$, by induction. Base case: $G^{(0)} = G = \gamma_1(G)$. Suppose $G^{(i)} \le \gamma_{i+1}(G)$. Then

using $G^{(i)} \le G$ and $G^{(i)} \le \gamma_{i+1}(G)$ and monotonicity of the commutator bracket in each argument. If $G$ is nilpotent, $\gamma_{c+1}(G) = \{e\}$ for some $c$, so $G^{(c)} \le \gamma_{c+1}(G) = \{e\}$, i.e. $G^{(c)} = \{e\}$: $G$ is solvable.

(c) Finite $p$-groups are nilpotent. Let $|G| = p^n$. We show $\gamma_{i+1}(G) \ne \gamma_i(G)$ whenever $\gamma_i(G) \ne \{e\}$, which forces the strictly descending chain $G = \gamma_1(G) \supsetneq \gamma_2(G) \supsetneq \cdots$ to reach $\{e\}$ in at most $n$ steps (a strictly decreasing chain of subgroups of a group of order $p^n$ has length at most $n$, since each proper containment of subgroups of a $p$-group drops the order by at least a factor of $p$). Concretely we use the upper central series instead, which is cleaner: since $G$ is a nontrivial finite $p$-group, $Z(G) = Z_1(G) \ne \{e\}$ (standard fact for $p$-groups). Consider $G/Z_1(G)$; if nontrivial it is again a $p$-group (a quotient of a $p$-group by a normal subgroup has order a power of $p$), so it too has nontrivial center, meaning $Z_2(G)/Z_1(G) = Z(G/Z_1(G)) \ne \{e\}$, so $Z_2(G) \supsetneq Z_1(G)$. Inductively, as long as $Z_i(G) \ne G$, the quotient $G/Z_i(G)$ is a nontrivial finite $p$-group (order $p^n/|Z_i(G)|$, still $>1$), hence has nontrivial center, forcing $Z_{i+1}(G) \supsetneq Z_i(G)$. Since $|Z_i(G)|$ strictly increases and is bounded by $p^n$, we must reach $Z_c(G) = G$ for some $c \le n$. By part (a), $G$ is nilpotent.

(d) Structure theorem for finite nilpotent groups. Lemma (normalizers grow in nilpotent groups): If $G$ is nilpotent and $H \lneq G$ is a proper subgroup, then $H \lneq N_G(H)$ (i.e. $H$ is not self-normalizing). Proof of lemma: Since $G$ is nilpotent, $Z_0(G) = \{e\} \le H$ and $Z_c(G) = G \not\le H$ for the nilpotency class $c$. Let $i$ be the largest index with $Z_i(G) \le H$; then $i < c$ and $Z_{i+1}(G) \not\le H$. Pick $z \in Z_{i+1}(G) \setminus H$. For any $h \in H$, since $zZ_i(G) \in Z_{i+1}(G)/Z_i(G) = Z(G/Z_i(G))$, the coset $zZ_i(G)$ commutes with $hZ_i(G)$ in $G/Z_i(G)$, i.e. $[z,h] \in Z_i(G) \le H$. So $z^{-1}h^{-1}zh \in H$, hence $h^{-1}zh = z(z^{-1}h^{-1}zh) \in zH$. Also $hzh^{-1}$: from $z^{-1}h^{-1}zh \in H$ we get $zh z^{-1} \cdot h^{-1} \cdot$(rearranging) — more directly: $[z,h] \in H$ means $z^{-1}h^{-1}zh = h_0 \in H$, so $zh = h(zh_0)$... let us just directly verify $zHz^{-1} = H$: for $h \in H$, $zhz^{-1} = h \cdot h^{-1}zhz^{-1} = h \cdot [h^{-1}, z^{-1}]^{-1}$-type manipulation. Cleanest: $z^{-1}h^{-1}zh \in Z_i(G) \le H$ means $h^{-1}zh \in zH$, i.e. $h^{-1}zh = zh'$ for some $h' \in H$; rearranging, $zh = hzh'$, so $z = hzh'h^{-1} \in$ ... We simplify: from $z^{-1}h^{-1}zh \in H$ for every $h \in H$, set $k = z^{-1}h^{-1}zh \in H$; then $z^{-1}h^{-1}z = kh^{-1} \in H$, so $z^{-1}Hz \subseteq H$ hence (as $H$ is finite, this inclusion of equal-order sets is equality, or directly apply the same argument to $h^{-1}$ ranging over $H$) $z^{-1}Hz = H$, i.e. $z \in N_G(H)$. Since $z \notin H$, $N_G(H) \supsetneq H$. This proves the lemma. Now suppose $G$ finite nilpotent, and let $P$ be a Sylow $p$-subgroup of $G$. We claim $P \trianglelefteq G$. Let $N = N_G(P)$. A standard Sylow-theory fact is $N_G(N_G(P)) = N_G(P)$ (the "normalizer of the normalizer" is itself, since $P$ is the unique Sylow $p$-subgroup of $N_G(P)$ and any $g$ normalizing $N_G(P)$ must send $P$ to a Sylow $p$-subgroup of $N_G(P)$, hence to $P$ itself, so $g \in N_G(P)$). If $N \ne G$, the lemma gives $N \lneq N_G(N) = N$, a contradiction. So $N = G$, i.e. $P \trianglelefteq G$. So every Sylow subgroup of a finite nilpotent group is normal, giving the "only if" of the second equivalence in (d). Conversely, suppose every Sylow subgroup of $G$ is normal. Let $p_1, \ldots, p_k$ be the distinct primes dividing $|G|$, and $P_1,\ldots,P_k$ the corresponding (unique, since normal Sylow subgroups are unique — any two Sylow $p$-subgroups are conjugate, and a normal subgroup is its own only conjugate) Sylow subgroups. For $i \ne j$, $P_i \cap P_j = \{e\}$ since their orders are coprime prime powers (Lagrange: the order of any element of the intersection divides both $|P_i|$ and $|P_j|$, which are coprime). Since each $P_i \trianglelefteq G$, for $x \in P_i, y \in P_j$ ($i\ne j$) the commutator $[x,y] = x^{-1}(y^{-1}xy) = (x^{-1}y^{-1}x)y$ lies in $P_i \cap P_j = \{e\}$ using normality of both (the first parenthesization shows $[x,y] \in P_i$ since $y^{-1}xy \in P_i$ by normality of $P_i$, wait we need $x^{-1}\cdot(y^{-1}xy) \in P_i$ — yes since $x^{-1} \in P_i$ and $y^{-1}xy \in P_i$ as $P_i \trianglelefteq G$; the second parenthesization shows $[x,y] \in P_j$ similarly). So $[x,y] = e$, i.e. elements of distinct Sylow subgroups commute. The map $\varphi: P_1 \times \cdots \times P_k \to G$, $(x_1,\ldots,x_k) \mapsto x_1 x_2 \cdots x_k$, is then a homomorphism (using that distinct $P_i$'s commute elementwise to check $\varphi((x_i)(y_i)) = \varphi((x_i))\varphi((y_i))$ directly). It is injective: if $x_1\cdots x_k = e$, then for each prime $p_i$, isolating $x_i = (x_1\cdots x_{i-1})^{-1}(x_{i+1}\cdots x_k)^{-1}$ shows $x_i$ lies in the subgroup generated by the other (pairwise commuting, coprime-order) $P_j$'s, whose order is coprime to $p_i$; since $x_i \in P_i$ has order a power of $p_i$, and also order dividing that coprime number, $x_i = e$. It is surjective since $|P_1\times\cdots\times P_k| = \prod |P_i| = |G|$ by Sylow's theorem (the order of $G$ factors as the product of the orders of its Sylow subgroups), and an injective map between finite sets of equal size is a bijection. So $G \cong P_1 \times \cdots \times P_k$. Finally, each finite $p$-group $P_i$ is nilpotent by (c), and a finite direct product of nilpotent groups is nilpotent (one checks directly that $\gamma_j(H_1 \times \cdots \times H_k) = \gamma_j(H_1) \times \cdots \times \gamma_j(H_k)$ by induction on $j$, since commutators in a direct product are computed coordinatewise; hence if each $\gamma_{c_i+1}(H_i) = \{e\}$, taking $c = \max c_i$ gives $\gamma_{c+1}(\prod H_i) = \{e\}$). So $G \cong \prod P_i$ is nilpotent. This completes (d). $\blacksquare$