---
title: Classifying Groups of Small Order
module: Products and Group Structure
moduleNumber: 5
lessonNumber: 4
order: 504
summary: >
  With Sylow's theorem to force normal subgroups, direct and semidirect products
  to assemble them, and presentations to name the result, every group up to order
  fifteen can be listed explicitly. Free groups make presentations precise:
  generators with no relations, from which any group is a quotient by the normal
  closure of its relations.
topics: [Products and Group Structure]
sources:
  - book: Dummit & Foote
    ref: "Ch. 6 — Further Topics; §6.2 Applications in Groups of Medium Order, §6.3 A Word on Free Groups; Ch. 5 §5.3 Table of Groups of Small Order"
draft: false
---

Classifying the groups of a fixed order $n$ is the concrete form of the Hölder
program: find the simple building blocks, then find how they assemble. Three
tools do the work. [Sylow's
theorem](/abstract-algebra/group-actions-and-sylow/sylow-theorems) produces
normal subgroups; [direct](/abstract-algebra/products-and-group-structure/direct-products-and-finite-abelian-groups)
and [semidirect](/abstract-algebra/products-and-group-structure/semidirect-products)
products glue them together; and **presentations** name the outcome by generators
and relations. Free groups make that last notion precise.

## The classification strategy

For most small $n$ the plan is uniform.[^df-strategy] The order factors as $n =
p_1^{a_1} \cdots p_k^{a_k}$, and the goal is to force some Sylow subgroup to be
normal, then peel the group apart.

```algorithm
caption: $\textsc{Classify}(n)$ — list groups of order $n$
factor $n = p_1^{a_1} \cdots p_k^{a_k}$
for each prime $p$ dividing $n$ do
  compute the permissible Sylow counts $n_p$ with $n_p \equiv 1 \pmod p$ and $n_p \mid m$
find a normal subgroup $H$ // often a Sylow subgroup with $n_p = 1$
let $K$ be a complement, so $G = H K$ and $H \cap K = 1$
for each pair $(H, K)$ and each $\varphi : K \to \operatorname{Aut}(H)$ do
  form $H \rtimes_\varphi K$
deduplicate the semidirect products up to isomorphism
return the distinct isomorphism types
```

The hard step is forcing a normal subgroup. When the Sylow count $n_p$ is
pinned to $1$ by the congruence and divisibility constraints, the Sylow
$p$-subgroup is unique, hence normal, and gives the factor $H$. The number of
gluings $\varphi$ is then small.

$$
% caption: A first split of small orders by their prime factorization. Prime
% orders force cyclic groups; prime-square orders are abelian; orders $pq$ split
% on whether $p \mid q-1$; higher orders need the full Sylow-and-product machine.
\begin{tikzpicture}[scale=1.0, font=\small,
  nd/.style={draw, black, inner sep=3pt, fill=acc!8, align=center},
  lf/.style={draw, acc, inner sep=3pt, fill=acc!14, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[nd] (root) at (0,3.2) {order $n$};
\node[nd] (p)   at (-4.2,1.7) {$n = p$};
\node[nd] (p2)  at (-1.5,1.7) {$n = p^2$};
\node[nd] (pq)  at (1.5,1.7) {$n = pq$};
\node[nd] (big) at (4.4,1.7) {$n$ composite};
\node[lf] (pl)  at (-4.2,0.2) {$\mathbb{Z}_p$};
\node[lf] (p2l) at (-1.5,0.2) {$\mathbb{Z}_{p^2}$ or\\$(\mathbb{Z}_p)^2$};
\node[lf] (pql) at (1.5,0.2) {$\mathbb{Z}_{pq}$;\\sometimes one more};
\node[lf] (bigl) at (4.4,0.2) {Sylow $+$\\products};
\draw[black] (root) -- (p);
\draw[black] (root) -- (p2);
\draw[black] (root) -- (pq);
\draw[black] (root) -- (big);
\draw[black, dashed] (p) -- (pl);
\draw[black, dashed] (p2) -- (p2l);
\draw[black, dashed] (pq) -- (pql);
\draw[black, dashed] (big) -- (bigl);
\end{tikzpicture}
$$

## The base cases

Three families of orders are settled by earlier results, and they cover most
small $n$.

- **Order $p$.** Every group of prime order is cyclic, $\mathbb{Z}_p$, by
  Lagrange: a nonidentity element has order dividing $p$, hence order $p$, so it
  generates.
- **Order $p^2$.** Every group of order $p^2$ is abelian, since a $p$-group has
  nontrivial center and $G/Z(G)$ cyclic forces $G$ abelian. The two abelian
  types are $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p \times \mathbb{Z}_p$.
- **Order $pq$** ($p < q$ primes). The Sylow $q$-subgroup is normal, so $G \cong
  \mathbb{Z}_q \rtimes \mathbb{Z}_p$. If $p \nmid q-1$ the only group is cyclic
  $\mathbb{Z}_{pq}$; if $p \mid q-1$ there is one additional non-abelian group.

> **Worked example.** Every group of order $15$ is cyclic. Factor $15 = 3 \cdot
> 5$. For the Sylow $5$-subgroup, $n_5 \mid 3$ and $n_5 \equiv 1 \pmod 5$ force
> $n_5 = 1$; for the Sylow $3$-subgroup, $n_3 \mid 5$ and $n_3 \equiv 1 \pmod 3$
> force $n_3 = 1$. Both Sylow subgroups are normal, of coprime orders $5$ and
> $3$, so they intersect trivially and commute, and the recognition theorem gives
> $$
> G \cong \mathbb{Z}_3 \times \mathbb{Z}_5 \cong \mathbb{Z}_{15}.
> $$
> This is the case $p \nmid q-1$ of the order-$pq$ count, with $p = 3$, $q = 5$,
> and $3 \nmid 4$.

Order $p^3$ adds the first non-abelian $p$-group examples. For $p$ odd there are
exactly two non-abelian groups of order $p^3$, distinguished by whether they
contain an element of order $p^2$; for $p = 2$ these are $D_8$ and $Q_8$. The
abelian ones follow the [partition
count](/abstract-algebra/products-and-group-structure/direct-products-and-finite-abelian-groups):
three of them, from the partitions $3$; $2,1$; $1,1,1$.

## Forcing non-simplicity

When no Sylow count is immediately $1$, the classification depends on ruling out
simple groups of order $n$ — showing every such group has a proper nontrivial
normal subgroup. Three counting techniques recur.[^df-techniques]

**Counting elements of prime order.** If a Sylow $p$-subgroup has order $p$, then
distinct conjugates intersect trivially, so the $n_p$ conjugates contribute
$n_p(p-1)$ elements of order $p$. Summing these counts over several primes can
exceed $|G|$, which is impossible, forcing some $n_p = 1$.

> **Worked example.** No simple group of order $105 = 3 \cdot 5 \cdot 7$. Suppose
> $G$ were simple, so no Sylow count equals $1$. The constraints $n_p \equiv 1
> \pmod p$ and $n_p \mid m$ force $n_3 = 7$, $n_5 = 21$, and $n_7 = 15$. Distinct
> Sylow subgroups of prime order meet trivially, so counting elements of each
> prime order,
> $$
> n_3(3-1) + n_5(5-1) + n_7(7-1) = 14 + 84 + 90 = 188 > 105,
> $$
> more elements than the group holds. The contradiction forces some $n_p = 1$, so
> $G$ has a normal Sylow subgroup and is not simple.

**Exploiting subgroups of small index.** A subgroup $H$ of index $k$ gives an
action of $G$ on the $k$ cosets, hence a homomorphism $G \to S_k$ whose kernel
lies in $H$. If $G$ is simple the kernel is trivial, so $G$ embeds in $S_k$ and
$|G| \mid k!$. This bounds how small an index a simple group can have and often
contradicts a Sylow count.

**Permutation representations.** Refining the previous idea, once $G \leq S_k$
one compares Sylow subgroups inside $S_k$. If $P \in \operatorname{Syl}_p(G)$ is
also Sylow in $S_k$, then $|N_G(P)|$ must divide $|N_{S_k}(P)|$, a numerical
constraint that frequently fails.

$$
% caption: Three routes to a normal subgroup when Sylow counts are not
% immediately forced. Each derives a contradiction from the assumption that $G$
% is simple, so some $n_p$ collapses to $1$.
\begin{tikzpicture}[scale=1.0, font=\small,
  nd/.style={draw, black, inner sep=4pt, fill=acc!8, align=center, minimum width=30mm}]
\definecolor{acc}{HTML}{4A6FA5}
\node[nd, fill=acc!14, draw=acc] (assume) at (0,2.4) {assume $G$ simple};
\node[nd] (count) at (-3.6,0.9) {count prime-order\\elements};
\node[nd] (index) at (0,0.9) {small-index\\coset action};
\node[nd] (perm)  at (3.6,0.9) {Sylow counts\\inside $S_k$};
\node[nd, fill=acc!14, draw=acc] (contra) at (0,-0.7) {contradiction: some $n_p = 1$};
\draw[black, ->] (assume) -- (count);
\draw[black, ->] (assume) -- (index);
\draw[black, ->] (assume) -- (perm);
\draw[acc, thick, ->] (count) -- (contra);
\draw[acc, thick, ->] (index) -- (contra);
\draw[acc, thick, ->] (perm) -- (contra);
\end{tikzpicture}
$$

These methods show, for example, that there are no simple groups of order $56$,
$105$, or $396$. Not every order yields to them — genuine simple groups exist at
$60$, $168$, $360$, and beyond — but for the small orders below, they always
suffice to find a normal subgroup and begin the assembly.

## The table of small orders

Combining the base cases with Sylow-driven splitting produces the complete
census through order $15$, with order $16$ recorded for scale.[^df-table]

| $n$ | count | abelian | non-abelian |
| --- | --- | --- | --- |
| $1$ | $1$ | trivial | — |
| $2$ | $1$ | $\mathbb{Z}_2$ | — |
| $3$ | $1$ | $\mathbb{Z}_3$ | — |
| $4$ | $2$ | $\mathbb{Z}_4$, $\mathbb{Z}_2 \times \mathbb{Z}_2$ | — |
| $5$ | $1$ | $\mathbb{Z}_5$ | — |
| $6$ | $2$ | $\mathbb{Z}_6$ | $S_3 \cong D_6$ |
| $7$ | $1$ | $\mathbb{Z}_7$ | — |
| $8$ | $5$ | $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$, $\mathbb{Z}_2^{\,3}$ | $D_8$, $Q_8$ |
| $9$ | $2$ | $\mathbb{Z}_9$, $\mathbb{Z}_3 \times \mathbb{Z}_3$ | — |
| $10$ | $2$ | $\mathbb{Z}_{10}$ | $D_{10}$ |
| $11$ | $1$ | $\mathbb{Z}_{11}$ | — |
| $12$ | $5$ | $\mathbb{Z}_{12}$, $\mathbb{Z}_6 \times \mathbb{Z}_2$ | $A_4$, $D_{12}$, $\mathbb{Z}_3 \rtimes \mathbb{Z}_4$ |
| $13$ | $1$ | $\mathbb{Z}_{13}$ | — |
| $14$ | $2$ | $\mathbb{Z}_{14}$ | $D_{14}$ |
| $15$ | $1$ | $\mathbb{Z}_{15}$ | — |
| $16$ | $14$ | five types | nine types |

Prime orders and the order $15$ (where $3 \nmid 5-1$) give a single cyclic
group. Prime-square orders give two abelian types. Non-abelian groups first
appear at order $6$, and the count jumps whenever the order carries a large
prime power: order $16$ already has fourteen types, and the $p$-group tables
grow rapidly from there.

## Free groups

Naming these groups needs a precise notion of "generators and relations." A
presentation like $D_{2n} = \langle r, s \mid r^n = s^2 = 1,\ s^{-1}rs =
r^{-1}\rangle$ has to mean something definite. The **free group** supplies the
raw material: generators subject to no relations at all.

> **Definition (Free group).** For a set $S$, the **free group** $F(S)$ is the
> set of **reduced words** in the symbols of $S$ and their formal inverses —
> finite strings with no adjacent cancelling pair $x x^{-1}$ — under
> concatenation followed by reduction. The cardinality of $S$ is the **rank** of
> $F(S)$.

In $F(\{a,b\})$ the elements are strings like $a$, $ab$, $abab$, $ba$, and
$aba^{-1}b^2$, all distinct: no relation forces any two reduced words to
coincide. Concatenation multiplies them, cancelling only juxtaposed inverses, so
$(aba)(a^{-1}b) = ab^2$. The Cayley graph of $F(\{a,b\})$ is the infinite
$4$-valent tree — no loops, because a loop would be a nontrivial relation.

$$
% caption: Part of the Cayley graph of the free group on $a, b$, the infinite
% $4$-regular tree; here $A = a^{-1}$ and $B = b^{-1}$. Every reduced word is a
% distinct vertex, and the absence of cycles is the absence of relations.
\begin{tikzpicture}[scale=1.0, font=\small,
  v/.style={circle, draw, black, fill=acc!10, inner sep=1.3pt}]
\definecolor{acc}{HTML}{4A6FA5}
\node[v, fill=acc!18, draw=acc] (o) at (0,0) {$1$};
\node[v] (a) at (2.2,0)  {$a$};
\node[v] (A) at (-2.2,0) {$A$};
\node[v] (b) at (0,2.0)  {$b$};
\node[v] (B) at (0,-2.0) {$B$};
\draw[black] (o) -- (a) node[midway, above, black, font=\scriptsize] {$a$};
\draw[black] (o) -- (A) node[midway, above, black, font=\scriptsize] {$A$};
\draw[black] (o) -- (b) node[midway, right, black, font=\scriptsize] {$b$};
\draw[black] (o) -- (B) node[midway, right, black, font=\scriptsize] {$B$};
% second level off a
\node[v] (aa) at (3.7,0.9)  {$aa$};
\node[v] (ab) at (3.7,-0.9) {$ab$};
\draw[black] (a) -- (aa);
\draw[black] (a) -- (ab);
% second level off b
\node[v] (ba) at (1.3,3.0)  {$ba$};
\node[v] (bb) at (-1.3,3.0) {$bb$};
\draw[black] (b) -- (ba);
\draw[black] (b) -- (bb);
\end{tikzpicture}
$$

The defining feature of freeness is a mapping property: because the generators
satisfy no relations, they may be sent anywhere.

> **Theorem (Universal property).** For any group $G$ and any set map $\varphi :
> S \to G$, there is a unique homomorphism $\Phi : F(S) \to G$ extending
> $\varphi$. Thus $F(S)$ is determined up to unique isomorphism by $S$.

A homomorphism out of $F(S)$ is fixed by naming the image of each generator, and
those images are unconstrained.[^df-universal] This is the exact analogue of a
basis of a vector space, where any assignment of the basis vectors extends
uniquely to a linear map; $F(S)$ is the "vector space with basis $S$" in the
category of groups.

## Presentations as quotients

Every group is a quotient of a free group. Taking $S$ to be a generating set of
$G$, the universal property gives a surjection $\pi : F(S) \to G$, and $G \cong
F(S)/\ker \pi$. A **presentation** records $G$ by the generators $S$ together
with enough relations to cut out the kernel.

> **Definition (Presentation).** A **presentation** $\langle S \mid R\rangle$ of
> $G$ consists of a generating set $S$ and a set $R$ of words in $F(S)$ such that
> the **normal closure** of $R$ — the smallest normal subgroup containing $R$ —
> equals $\ker\pi$. The elements of $S$ are **generators**, those of $R$ are
> **relations**.

The kernel is not $R$ itself but the subgroup generated by $R$ and all its
conjugates, since the kernel must be normal.[^df-presentation] Collapsing that
kernel imposes exactly the relations in $R$ and nothing more.

$$
% caption: A presentation realizes $G$ as a quotient of the free group $F(S)$:
% the surjection $\pi$ collapses the normal closure of the relations $R$, and
% $G \cong F(S)/\ker\pi$ where $\ker\pi$ is that normal closure.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% free group blob
\draw[black, fill=acc!7] (0,0) ellipse (2.0 and 1.4);
\node[black] at (0,0.85) {$F(S)$};
% kernel inside
\draw[acc, fill=acc!16] (0,-0.35) ellipse (1.1 and 0.6);
\node[acc, font=\scriptsize] at (0,-0.35) {normal closure of $R$};
% target blob
\draw[black, fill=acc!10] (6.0,0) ellipse (1.5 and 1.0);
\node[black] at (6.0,0) {$G$};
% surjection
\draw[acc, thick, ->] (2.1,0.3) -- (4.4,0.3) node[midway, above, black, font=\scriptsize] {quotient map};
\node[black, font=\scriptsize, anchor=north] at (3.2,0.15) {collapses the kernel};
\end{tikzpicture}
$$

To verify that $\langle S \mid R\rangle$ presents a known finite group $G$, two
checks suffice: $S$ generates $G$, and any group generated by $S$ subject to $R$
has order at most $|G|$. For $D_{2n} = \langle r, s \mid r^n = s^2 = 1,\ s^{-1}rs
= r^{-1}\rangle$, the relations imply a normal subgroup $\langle r\rangle$ of
order at most $n$ with quotient generated by $s$ of order at most $2$, so any
such group has order at most $2n$; since $D_{2n}$ of order $2n$ satisfies the
relations, the presentation is exactly $D_{2n}$.[^df-presentation]

Familiar groups get compact presentations:

| Group | Presentation |
| --- | --- |
| $\mathbb{Z}$ | $\langle a \rangle$ |
| $\mathbb{Z} \times \mathbb{Z}$ | $\langle a, b \mid ab = ba\rangle$ |
| $\mathbb{Z}_n \times \mathbb{Z}_m$ | $\langle a, b \mid a^n = b^m = 1,\ ab = ba\rangle$ |
| $D_{2n}$ | $\langle r, s \mid r^n = s^2 = 1,\ s^{-1}rs = r^{-1}\rangle$ |
| $Q_8$ | $\langle i, j \mid i^4 = 1,\ j^2 = i^2,\ j^{-1}ij = i^{-1}\rangle$ |

Presentations are compact but hard to decode: in general it is undecidable
whether a given set of generators and relations defines the trivial group or an
infinite one, so a compact presentation may hide a group that is hard to
identify.[^df-hard]

## The reach and limit of the program

Assembling Sylow's theorem, products, and presentations classifies every group
through order $15$ and, with more labor, well beyond. The method is inductive:
each group breaks into a normal subgroup and a complement, both smaller and
already understood. It stalls at the simple groups, which admit no such
decomposition; their classification is a separate and far deeper project, one
strand of which appears in the [simplicity of
$A_n$](/abstract-algebra/group-actions-and-sylow/automorphisms-and-simple-groups).

[^df-strategy]: **Dummit & Foote**, _Abstract Algebra_, §6.2 — Applications in Groups of Medium Order: the general approach of computing permissible Sylow numbers and forcing a normal subgroup, motivated by the Hölder program.
[^df-techniques]: **Dummit & Foote**, §6.2: the techniques of counting elements of prime order, exploiting subgroups of small index via the coset action $G \to S_k$, and comparing Sylow subgroups within $S_k$; the order-$105$ and order-$56$ examples.
[^df-table]: **Dummit & Foote**, §5.3 — Table of Groups of Small Order: the isomorphism types through order $15$ and the count of fourteen groups of order $16$.
[^df-universal]: **Dummit & Foote**, §6.3 — A Word on Free Groups, Theorems 16 and 17: the construction of $F(S)$ from reduced words and its universal property, with the vector-space basis analogy.
[^df-presentation]: **Dummit & Foote**, §6.3: presentations as quotients $F(S)/N$ where $N$ is the normal closure of the relations, and the order-bound test for verifying a presentation of $D_{2n}$.
[^df-hard]: **Dummit & Foote**, §6.3, Example 4: it is in general extremely difficult to determine from a presentation whether the group is trivial, finite, or infinite.