01.02.05 · foundations / groups

Solvable group, nilpotent group, Jordan-Holder theorem

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Anchor (Master): Jordan 1869; Holder 1889 *Math. Ann.* 34; Burnside 1897 *Proc. London Math. Soc.* 28; Dummit-Foote §3.4, §6.1; Rotman *An Introduction to the Theory of Groups*

Intuition [Beginner]

Some groups can be taken apart layer by layer until nothing is left but simple pieces. A solvable group admits this dissection. Each layer is an abelian quotient — the pieces at that level commute — and peeling them off one at a time eventually reduces the group to the identity.

Think of factoring the number $60$ into primes: $60 = 2 \times 2 \times 3 \times 5$ . A solvable group factors in a similar way, but the "prime pieces" are simple groups, and the factoring process uses quotients instead of division. The symmetric group on five or more symbols cannot be broken down this way.

A nilpotent group is even more structured. In a nilpotent group, elements that sit far enough apart in the group's internal hierarchy commute with each other. Every abelian group is nilpotent, and every nilpotent group is solvable, but not conversely.

Why does this concept exist? Solvability and nilpotence classify groups by how close they are to being abelian, and the Jordan-Holder theorem guarantees the simple pieces in this decomposition are unique.

Visual [Beginner]

The diagram shows a tower for $S_{3}$ . At the top is the full group of six permutations. One level down sits the subgroup $A_{3}$ of even permutations with three elements. At the base sits the identity element alone. Each step in the tower produces an abelian quotient.

The two quotient layers have sizes $2$ and $3$ . Each is a simple abelian group — a cyclic group of prime order. The Jordan-Holder theorem says these layers are unique: any other way of breaking $S_{3}$ into simple pieces produces the same sizes.

Worked example [Beginner]

Consider $S_{3}$ , the group of permutations of three objects, with $6$ elements.

Step 1. Inside $S_{3}$ sits $A_{3}$ , the alternating group of even permutations, with $3$ elements. The quotient $S_{3} / A_{3}$ has $2$ elements and is abelian.

Step 2. Inside $A_{3}$ , the only proper subgroup is the one containing just the identity element. The quotient $A_{3}$ modulo this subgroup has $3$ elements and is also abelian.

Step 3. The chain $S_{3} \supset A_{3} \supset {e}$ reaches the identity in two steps. The quotient layers have sizes $2$ and $3$ , both prime, hence both simple.

What this tells us: $S_{3}$ is solvable because it breaks into abelian layers, and the two layers (of sizes $2$ and $3$ ) are unique.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $G$ be a group. The derived series of $G$ is defined inductively:

$G^{(0)} = G, G^{(i + 1)} = [G^{(i)}, G^{(i)}]$

where $[H, H]$ denotes the commutator subgroup generated by all elements $hk h^{- 1} k^{- 1}$ with $h, k \in H$ . A group $G$ is solvable if $G^{(n)} = {e}$ for some $n \geq 0$ . The smallest such $n$ is the derived length of $G$ .

A composition series for a group $G$ is a chain of subgroups

${e} = G_{0} ⊴ G_{1} ⊴ \dots ⊴ G_{r} = G$

such that each $G_{i - 1}$ is normal in $G_{i}$ and each quotient $G_{i} / G_{i - 1}$ is a simple group (a group with no proper non-identity normal subgroups). The quotients $G_{i} / G_{i - 1}$ are the composition factors. Every finite group admits a composition series; existence follows by refining any subnormal series, using the fact that a finite group has maximal normal subgroups at each step. The length $r$ is the composition length ^{[Dummit-Foote §3.4]}.

The upper central series of $G$ is $Z_{0} = {e}$ , $Z_{i + 1} / Z_{i} = Z (G / Z_{i})$ , where $Z (G / Z_{i})$ is the center of the quotient. A group $G$ is nilpotent if $Z_{c} = G$ for some $c \geq 0$ ; the smallest such $c$ is the nilpotence class. Equivalently, $G$ is nilpotent of class at most $c$ if and only if every iterated commutator $[g_{1}, [g_{2}, [\dots, [g_{c}, g_{c + 1}] \dots]]]$ equals the identity for all choices of elements ^{[Rotman Ch. 5]}.

Counterexamples to common slips [Intermediate+]

Solvability does not imply nilpotence. The group $S_{3}$ is solvable (derived length $2$ ) but not nilpotent: its upper central series satisfies $Z_{0} = {e}$ and $Z_{1} = Z (S_{3}) = {e}$ (the center of $S_{3}$ is the identity), so the series never reaches $S_{3}$ .
Composition series need not equal the derived series. The cyclic group $Z /6 Z$ has derived length $1$ (it is abelian) but composition length $2$ , since $Z /6 Z ≅ Z /2 Z \times Z /3 Z$ and the composition factors are $Z /2 Z$ and $Z /3 Z$ .

Key theorem with proof [Intermediate+]

Theorem (Jordan-Holder). Let $G$ be a finite group with two composition series

${e} = G_{0} ⊴ G_{1} ⊴ \dots ⊴ G_{r} = G$

and

${e} = H_{0} ⊴ H_{1} ⊴ \dots ⊴ H_{s} = G .$

Then $r = s$ , and the multisets of composition factors ${G_{i} / G_{i - 1}}_{i = 1}^{r}$ and ${H_{j} / H_{j - 1}}_{j = 1}^{s}$ are equal up to isomorphism and reordering.

Proof. The proof proceeds by strong induction on $min (r, s)$ .

Base case ( $r = 1$ ). If $r = 1$ , then $G$ is simple. The only composition series of a simple group is ${e} ⊴ G$ , so $s = 1$ and the result holds.

Inductive step. Assume the theorem holds for all groups possessing a composition series of length strictly less than $min (r, s)$ . Consider two composition series of $G$ of lengths $r$ and $s$ as above.

Case 1: $G_{r - 1} = H_{s - 1}$ . Set $N = G_{r - 1} = H_{s - 1}$ . Both series restrict to composition series of $N$ , of lengths $r - 1$ and $s - 1$ . By the induction hypothesis applied to $N$ , the lengths agree ( $r - 1 = s - 1$ , so $r = s$ ) and the composition factors of $N$ coincide in both series. Both full series append the factor $G / N$ , so the composition factors of $G$ are the same.

Case 2: $G_{r - 1} \neq = H_{s - 1}$ . Set $M = G_{r - 1} \cap H_{s - 1}$ . Both $G_{r - 1}$ and $H_{s - 1}$ are normal in $G$ , so $M$ is normal in $G$ .

We first show $G_{r - 1} H_{s - 1} = G$ . Since $G_{r - 1} ⊴ G$ and $H_{s - 1} ⊴ G$ , the product $G_{r - 1} H_{s - 1}$ is a normal subgroup of $G$ containing $G_{r - 1}$ . The quotient $G_{r - 1} H_{s - 1} / G_{r - 1}$ is a normal subgroup of $G / G_{r - 1}$ . Since $G / G_{r - 1}$ is simple and $H_{s - 1} \neq \subseteq G_{r - 1}$ (because $H_{s - 1} \neq = G_{r - 1}$ ), this quotient is non-identity, hence equals $G / G_{r - 1}$ . Therefore $G_{r - 1} H_{s - 1} = G$ .

By the second isomorphism theorem:

$G_{r - 1} / M = G_{r - 1} / (G_{r - 1} \cap H_{s - 1}) ≅ G_{r - 1} H_{s - 1} / H_{s - 1} = G / H_{s - 1}$

which is simple. Similarly:

$H_{s - 1} / M ≅ G / G_{r - 1}$

which is also simple.

Now choose a composition series of $M$ :

${e} = M_{0} ⊴ M_{1} ⊴ \dots ⊴ M_{t} = M .$

Splicing this with $G_{r - 1}$ yields a composition series of $G_{r - 1}$ of length $t + 1$ :

${e} = M_{0} ⊴ \dots ⊴ M_{t} = M ⊴ G_{r - 1}$

with factors $M_{1} / M_{0}, \dots, M_{t} / M_{t - 1}, G_{r - 1} / M$ . The original series restricts to a composition series of $G_{r - 1}$ of length $r - 1$ . By induction, $t + 1 = r - 1$ , giving $t = r - 2$ , and the factors of $G_{r - 1}$ are the same in both series.

Similarly, splicing the composition series of $M$ with $H_{s - 1}$ gives a composition series of $H_{s - 1}$ of length $t + 1$ . By induction on $H_{s - 1}$ , we get $t + 1 = s - 1$ , giving $s = t + 2 = r$ .

The original series of $G$ has factors: those of $G_{r - 1}$ (which equal the factors of $M$ , plus $G_{r - 1} / M ≅ G / H_{s - 1}$ ), plus $G / G_{r - 1}$ . So the full list is: factors of $M$ , then $G / H_{s - 1}$ , then $G / G_{r - 1}$ .

The second series has factors: those of $H_{s - 1}$ (which equal the factors of $M$ , plus $H_{s - 1} / M ≅ G / G_{r - 1}$ ), plus $G / H_{s - 1}$ . So the full list is: factors of $M$ , then $G / G_{r - 1}$ , then $G / H_{s - 1}$ .

The two lists differ only in the order of the last two factors. $□$

Bridge. The Jordan-Holder theorem builds toward the classification of finite simple groups, where the composition factors form the irreducible atomic pieces of any finite group. This is exactly the group-theoretic analogue of prime factorization for integers. The foundational reason the theorem works is that any two chains of normal subgroups can be spliced at their intersection, and this identifies the composition factors of $G$ with an unordered multiset of simple groups. The bridge is between the lattice of subnormal chains and the multiset of simple quotients, and the pattern generalises to modules over a ring via the same refinement argument. Putting these together, the Jordan-Holder theorem appears again in 01.02.02 where the isomorphism theorems provide the quotient machinery, and in 01.02.04 where Sylow subgroups constrain the possible composition factors of a finite group.

Exercises [Intermediate+]

Exercise 7 (hard, symbolic).

Prove that $S_{n}$ is not solvable for $n \geq 5$ by showing that the derived series of $S_{n}$ stabilises at $A_{n}$ .

Hint

First show $[S_{n}, S_{n}] = A_{n}$ for all $n \geq 2$ . Then show $[A_{n}, A_{n}] = A_{n}$ for $n \geq 5$ by expressing every $3$ -cycle as a commutator of elements in $A_{n}$ .

Answer

For $n \geq 2$ , $[S_{n}, S_{n}] = A_{n}$ because $S_{n} / A_{n} ≅ Z /2 Z$ is abelian (so $[S_{n}, S_{n}] \subseteq A_{n}$ ) and $A_{n}$ is generated by $3$ -cycles, each of which is a commutator: $(i j k) = (i k j) (i j) (i k j)^{- 1} (i j)^{- 1}$ with $(i k j) \in A_{n}$ and $(i j) \in S_{n}$ . More directly, every $3$ -cycle is a commutator in $A_{n}$ for $n \geq 5$ : $(123) = [(14235), (13524)]$ using appropriate even permutations.

For $n \geq 5$ , $[A_{n}, A_{n}] = A_{n}$ because $A_{n}$ is simple (it has no proper non-identity normal subgroups) and $[A_{n}, A_{n}]$ is a normal subgroup of $A_{n}$ that is non-identity (it contains all $3$ -cycles). Since $A_{n}$ is simple, $[A_{n}, A_{n}] = A_{n}$ . The derived series is $S_{n} \supset A_{n} = A_{n} = \dots$ , which never reaches ${e}$ . Therefore $S_{n}$ is not solvable.

Exercise 8 (hard, symbolic).

Let $G$ be a finite nilpotent group. Prove that every Sylow subgroup of $G$ is normal. Deduce that $G$ is the direct product of its Sylow subgroups.

Hint

Use the fact that in a nilpotent group, every proper subgroup is properly contained in its normalizer. Apply this to a Sylow $p$ -subgroup $P$ and its normalizer $N_{G} (P)$ .

Answer

In a nilpotent group $G$ , every proper subgroup $H$ satisfies $H ⊊ N_{G} (H)$ (the normalizer of $H$ properly contains $H$ ). This is a standard property of nilpotent groups, provable via the upper central series: if $Z_{c} = G$ and $H ⊊ G$ , let $i$ be the largest index with $Z_{i} \subseteq H$ . Then $Z_{i + 1} \neq \subseteq H$ , and for $z \in Z_{i + 1} ∖ H$ and $h \in H$ , the commutator $[z, h] \in Z_{i} \subseteq H$ , so $z h z^{- 1} \in h Z_{i} \subseteq H$ . Thus $z \in N_{G} (H) ∖ H$ .

Let $P$ be a Sylow $p$ -subgroup of $G$ . Since $P$ is a Sylow subgroup, $N_{G} (P)$ is a subgroup containing $P$ . If $P \neq = N_{G} (P)$ , then $P ⊊ N_{G} (P)$ . Apply the property repeatedly: $P ⊊ N_{G} (P) ⊊ N_{G} (N_{G} (P)) ⊊ \dots$ . Since $G$ is finite, this chain terminates, and the only possibility is $N_{G} (P) = G$ , meaning $P ⊴ G$ .

Since each Sylow $p$ -subgroup $P_{p}$ is normal, $G$ contains the direct product $P_{p_{1}} \times \dots \times P_{p_{k}}$ for distinct primes $p_{1}, \dots, p_{k}$ dividing $∣ G ∣$ . The product has order $\prod ∣ P_{p_{i}} ∣ = ∣ G ∣$ , so $G = P_{p_{1}} \times \dots \times P_{p_{k}}$ .

Advanced results [Master]

Theorem 1 (Simplicity of $A_{n}$ , $n \geq 5$ ). The alternating group $A_{n}$ is simple for all $n \geq 5$ . The proof proceeds by showing that any non-identity normal subgroup of $A_{n}$ contains a $3$ -cycle, and the $3$ -cycles form a single conjugacy class in $A_{n}$ for $n \geq 5$ .

Theorem 2 (Non-solvability of $S_{n}$ , $n \geq 5$ , and Abel-Ruffini). The symmetric group $S_{n}$ is not solvable for $n \geq 5$ , because the derived series stabilises at $A_{n}$ and $[A_{n}, A_{n}] = A_{n}$ . The Abel-Ruffini theorem (Ruffini 1799, Abel 1824) states that the general polynomial of degree $n \geq 5$ is not solvable by radicals; the group-theoretic content is that the Galois group of a general polynomial of degree $n$ is $S_{n}$ , and radical solvability corresponds exactly to group solvability. ^{[Dummit-Foote §6.1]}

Theorem 3 (Feit-Thompson). Every finite group of odd order is solvable (Feit-Thompson 1963, Pacific J. Math. 13). The proof runs over 250 pages and is the foundational result that made the classification of finite simple groups possible.

Theorem 4 (Burnside $p^{a} q^{b}$ ). Every group of order $p^{a} q^{b}$ for primes $p, q$ is solvable (Burnside 1904). The proof uses character theory and is not purely group-theoretic. This appears again in 01.02.04 as a corollary of the Sylow theorems combined with representation theory.

Theorem 5 (Nilpotent = direct product of Sylow subgroups). A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups. Each Sylow $p$ -subgroup of a nilpotent group is normal. The forward direction uses the normalizer-growth property of nilpotent groups; the reverse uses the fact that every $p$ -group is nilpotent and direct products of nilpotent groups are nilpotent.

Theorem 6 (Lower central series). The lower central series $γ_{1} (G) = G$ , $γ_{i + 1} (G) = [γ_{i} (G), G]$ terminates at ${e}$ if and only if $G$ is nilpotent. A group is nilpotent of class $c$ if and only if $γ_{c + 1} (G) = {e}$ and $γ_{c} (G) \neq = {e}$ . The lower and upper central series give the same nilpotence class. ^{[Rotman Ch. 5]}

Theorem 7 (Closure properties). Subgroups, quotients, and finite direct products of solvable groups are solvable. Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent. If $N ⊴ G$ and both $N$ and $G / N$ are solvable, then $G$ is solvable. The corresponding statement for nilpotence fails: extensions of nilpotent groups need not be nilpotent.

Synthesis. The Jordan-Holder theorem is the foundational reason that every finite group decomposes uniquely into simple pieces, and this is exactly the group-theoretic analogue of prime factorization for integers. The central insight is that solvability constrains these simple pieces to be cyclic of prime order, which identifies a solvable group with a tower of prime-cyclic extensions. Putting these together with the Sylow theorems from 01.02.04, a nilpotent group is precisely one whose Sylow subgroups are normal and assemble as a direct product, and the bridge is between the commutator structure (measured by the derived and central series) and the prime-power decomposition (measured by Sylow subgroups). The pattern generalises from finite groups to profinite groups via inverse limits, to Lie algebras via the derived series of an ideal, and to modules over a ring via composition series in the Grothendieck group $K_{0}$ .

Full proof set [Master]

Proposition 1 (Lower central series characterises nilpotence). A group $G$ is nilpotent of class at most $c$ (i.e., $Z_{c} (G) = G$ ) if and only if $γ_{c + 1} (G) = {e}$ .

Proof. We prove both directions by induction on $c$ .

For the forward direction, we show $γ_{i + 1} (G) \subseteq Z_{c - i} (G)$ for all $0 \leq i \leq c$ , which gives $γ_{c + 1} (G) \subseteq Z_{0} (G) = {e}$ . The base case $i = 0$ : $γ_{1} (G) = G = Z_{c} (G)$ , which holds. For the inductive step, assume $γ_{i + 1} (G) \subseteq Z_{c - i} (G)$ . We need $γ_{i + 2} (G) = [γ_{i + 1} (G), G] \subseteq Z_{c - i - 1} (G)$ . Take $x \in γ_{i + 1} (G)$ and $g \in G$ . By hypothesis, $x \in Z_{c - i} (G)$ , meaning $x Z_{c - i - 1} (G)$ lies in $Z (G / Z_{c - i - 1} (G))$ . So $[x, g] \in Z_{c - i - 1} (G)$ . This establishes $γ_{i + 2} (G) \subseteq Z_{c - i - 1} (G)$ .

For the reverse direction, we show $Z_{i} (G) \cdot γ_{i + 1} (G) = G$ for all $i \geq 0$ . The base case $i = 0$ : $Z_{0} (G) \cdot γ_{1} (G) = {e} \cdot G = G$ . Inductive step: assume $Z_{i} (G) \cdot γ_{i + 1} (G) = G$ . We want $Z_{i + 1} (G) \cdot γ_{i + 2} (G) = G$ . Let $g \in G$ . Since $Z_{i} (G) \cdot γ_{i + 1} (G) = G$ , write $g = z x$ with $z \in Z_{i} (G)$ and $x \in γ_{i + 1} (G)$ . Then $g Z_{i} (G) = x Z_{i} (G)$ , and $[x, g^{'}] \in γ_{i + 2} (G)$ for all $g^{'} \in G$ . Since $γ_{c + 1} (G) = {e}$ , taking $i = c$ gives $Z_{c} (G) \cdot {e} = G$ , so $Z_{c} (G) = G$ . $□$

Proposition 2 (Subgroups and quotients of nilpotent groups are nilpotent). Let $G$ be nilpotent of class $c$ . Then every subgroup $H \leq G$ and every quotient $G / N$ is nilpotent of class at most $c$ .

Proof. We use the lower central series characterisation from Proposition 1. For a subgroup $H$ , we show $γ_{i} (H) \subseteq γ_{i} (G)$ by induction. The base case $γ_{1} (H) = H \subseteq G = γ_{1} (G)$ holds. Inductive step: $γ_{i + 1} (H) = [γ_{i} (H), H] \subseteq [γ_{i} (G), G] = γ_{i + 1} (G)$ . Since $γ_{c + 1} (G) = {e}$ , we get $γ_{c + 1} (H) \subseteq {e}$ , so $H$ is nilpotent of class at most $c$ .

For a quotient $G / N$ , observe that $γ_{i} (G / N) = γ_{i} (G) N / N$ (the lower central series commutes with quotients). This follows by induction: $γ_{1} (G / N) = G / N = γ_{1} (G) N / N$ , and $γ_{i + 1} (G / N) = [γ_{i} (G / N), G / N] = [γ_{i} (G) N / N, G / N] = [γ_{i} (G), G] N / N = γ_{i + 1} (G) N / N$ . Since $γ_{c + 1} (G) = {e}$ , we get $γ_{c + 1} (G / N) = N / N = {e N}$ , so $G / N$ is nilpotent of class at most $c$ . $□$

Connections [Master]

Group 01.02.01. The solvable and nilpotent hierarchies refine the basic group structure introduced in the group unit. A group is solvable when its commutator structure collapses to the identity through finitely many derived quotients; nilpotence strengthens this by requiring the collapse to happen via the center. The foundational reason these notions matter is that they stratify all finite groups by proximity to abelian groups.
Subgroup, coset, quotient group, isomorphism theorems 01.02.02. The Jordan-Holder theorem depends on the isomorphism theorems for its proof — specifically the second isomorphism theorem for splicing composition series at intersections, and the correspondence theorem for refining subnormal chains. The entire apparatus of normal subgroups and quotient groups from that unit is the prerequisite machinery for composition series.
Sylow theorems 01.02.04. The Sylow subgroups of a finite nilpotent group are all normal, and the group decomposes as their direct product. This is the strongest connection between Sylow theory and solvable structure: nilpotence is equivalent to simultaneous normality of all Sylow subgroups. The Sylow theorems also constrain the possible composition factors of a finite group, since each composition factor must be a simple group whose order divides the group order.

Historical & philosophical context [Master]

Camille Jordan 1869 established the uniqueness of composition factors for permutation groups in Commentaires sur Galois ^{[Jordan 1869]}, extending Galois's insight that certain groups decompose into simpler pieces through successive quotients. Otto Holder 1889 proved the full theorem for abstract groups in Zuruckfuhren einer algebraischen Gleichung auf eine Kette von Gleichungen (Math. Ann. 34) ^{[Holder 1889]}, introducing the modern language of composition series and factor groups. William Burnside 1897 initiated the systematic study of nilpotent groups in On the properties of groups whose order is a power of a prime (Proc. London Math. Soc. 28) ^{[Burnside 1897]}, recognizing that groups of prime-power order form a class with commutator-based structural properties distinct from general solvable groups.

The connection between solvability and the unsolvability of the quintic traces to Galois 1832. The Abel-Ruffini theorem (Ruffini 1799, Abel 1824) establishes that the general polynomial of degree five or higher cannot be solved by radicals. The group-theoretic content — that $S_{n}$ fails to be solvable for $n \geq 5$ — is the Galois-theoretic encoding of that result. Philip Hall 1928 developed Sylow systems in solvable groups, and the Feit-Thompson theorem (1963) confirmed that every finite group of odd order is solvable, the single deepest structural result in the classification of finite simple groups.

Bibliography [Master]

@article{Jordan1869,
  author = {Jordan, Camille},
  title = {Commentaires sur Galois},
  journal = {Mathematische Annalen},
  year = {1869},
}

@article{Holder1889,
  author = {H\"older, Otto},
  title = {Zur\"uckf\"uhren einer algebraischen Gleichung auf eine Kette von Gleichungen},
  journal = {Mathematische Annalen},
  volume = {34},
  year = {1889},
  pages = {26--56},
}

@article{Burnside1897,
  author = {Burnside, William},
  title = {On the properties of groups whose order is a power of a prime},
  journal = {Proceedings of the London Mathematical Society},
  volume = {28},
  year = {1897},
  pages = {1--28},
}

@book{DummitFoote,
  author = {Dummit, David S. and Foote, Richard M.},
  title = {Abstract Algebra},
  publisher = {John Wiley \& Sons},
  year = {2004},
  edition = {3rd},
}

@book{Rotman,
  author = {Rotman, Joseph J.},
  title = {An Introduction to the Theory of Groups},
  publisher = {Springer},
  year = {1995},
  edition = {4th},
}

@book{LangAlgebra,
  author = {Lang, Serge},
  title = {Algebra},
  publisher = {Springer},
  year = {2002},
  edition = {3rd},
}

Prerequisites

01.02.01

Tier anchors

beginner: breaking groups into simple layers
intermediate: Dummit-Foote §3.4, §6.1; Lang Algebra §I.3, §VI.7
master: Jordan 1869; Holder 1889 *Math. Ann.* 34; Burnside 1897 *Proc. London Math. Soc.* 28; Dummit-Foote §3.4, §6.1; Rotman *An Introduction to the Theory of Groups*

References

TODO_REF
Jordan — Commentaires sur Galois · 1869
TODO_REF
Holder — Zuruckfuhren einer algebraischen Gleichung auf eine Kette von Gleichungen · Math. Ann. 34, 1889
TODO_REF
Burnside — On properties of groups whose order is a power of a prime · Proc. London Math. Soc. 28, 1897
TODO_REF
Dummit-Foote — Abstract Algebra · §3.4, §6.1
TODO_REF
Rotman — An Introduction to the Theory of Groups · Ch. 5
TODO_REF
Lang — Algebra · §VI.7

Reviewer

TBD

Estimated time

beginner: 15m
intermediate: 40m
master: 75m