---
title: Subgroups and Their Substructures
module: Subgroups and Quotients
moduleNumber: 3
lessonNumber: 1
order: 301
summary: >
  A subgroup is a subset that is a group under the inherited operation. One
  test decides it: nonempty and closed under the map $(x,y) \mapsto xy^{-1}$.
  From an arbitrary subset $A$ we build the centralizer, normalizer, and center,
  and from an action the stabilizer and kernel, all of them subgroups nested in
  a fixed chain inside $G$.
topics: [Subgroups and Quotients]
sources:
  - book: Dummit & Foote
    ref: "Ch. 2 Subgroups; §2.1 Definition and Examples"
  - book: Dummit & Foote
    ref: "§2.2 Centralizers and Normalizers, Stabilizers and Kernels"
  - book: Judson
    ref: "Ch. 3; §3.3 Subgroups"
draft: false
---

A **subgroup** is a subset of a group that is itself a group under the same
operation, the group analogue of subspaces of a vector space, subrings of a
ring, or subfields of a field. Subgroups are cheap to recognize, and a handful
of them attach canonically to any subset of $G$ or any set $G$ acts on.

## Definition and shared identity

> **Definition (Subgroup).** A subset $H$ of a group $G$ is a **subgroup** of
> $G$ if $H$ is nonempty and closed under products and inverses: $x, y \in H$
> implies $xy \in H$ and $x^{-1} \in H$. We write $H \leq G$, and $H < G$ when
> the containment is proper.

The operation on $H$ is the operation of $G$ restricted to $H$; a subgroup is
never a subset carrying some unrelated multiplication. Because the operation is
inherited, every equation that holds in $H$ also holds in $G$, and the
cancellation laws in $G$ have two immediate consequences.[^df-def]

- **The identity is shared.** If $e_H$ is the identity of $H$, then
  $e_H e_H = e_H = e_H \cdot 1$ in $G$, so cancellation gives $e_H = 1$. Every
  subgroup contains the identity of $G$.
- **Inverses are shared.** The inverse of $x$ computed inside $H$ equals the
  inverse of $x$ computed inside $G$, so the notation $x^{-1}$ is unambiguous.

Two subgroups exist for free: $G$ itself and the **trivial subgroup**
$\{1\}$, written simply $1$. The relation $\leq$ is transitive: if $K \leq H$
and $H \leq G$ then $K \leq G$.

A few examples, drawing on the
[groups already built](/abstract-algebra/groups-and-symmetry/group-axioms-and-first-examples):

- $\mathbb{Z} \leq \mathbb{Q} \leq \mathbb{R}$ under addition; the even integers
  $2\mathbb{Z} \leq \mathbb{Z}$.
- In the [dihedral group](/abstract-algebra/groups-and-symmetry/dihedral-and-symmetric-groups)
  $D_{2n}$, the rotations $\{1, r, r^2, \dots, r^{n-1}\}$ form a subgroup of
  order $n$, since a product of rotations is a rotation and so is each inverse.
- The positive integers $\mathbb{Z}^+$ are **not** a subgroup of $\mathbb{Z}$:
  they miss the identity $0$ and are not closed under inverses.
- $\mathbb{Q} \setminus \{0\}$ under multiplication is not a subgroup of
  $\mathbb{R}$ under addition, even though both are groups and one is a subset of
  the other. The operations do not match.

## The subgroup criterion

Verifying the group axioms from scratch is tedious, mostly because of
associativity. Inside a known group that work is already done: associativity is
inherited, so a subset only needs closure. The two closure conditions collapse
into one.

> **Theorem (Subgroup Criterion).** A subset $H$ of a group $G$ is a subgroup
> if and only if $H \neq \varnothing$ and $xy^{-1} \in H$ for all $x, y \in H$.
> If $H$ is finite, it suffices that $H$ be nonempty and closed under products.

The proof is short.[^df-crit] Take any $x \in H$ (available
since $H$ is nonempty). Applying the condition to $y = x$ gives
$xx^{-1} = 1 \in H$. Applying it to the pair $1$ and $x$ gives
$1 \cdot x^{-1} = x^{-1} \in H$, so $H$ is closed under inverses. Finally, for
$x, y \in H$ we now have $y^{-1} \in H$, so $x(y^{-1})^{-1} = xy \in H$ and $H$
is closed under products. The finite refinement uses that in a finite group
every element has finite order: if $H$ is finite and closed under
multiplication, then for $x \in H$ the powers $x, x^2, x^3, \dots$ cannot all be
distinct, so $x^a = x^b$ for some $a < b$, whence $x^{b-a} = 1$ and
$x^{-1} = x^{b-a-1} \in H$.

$$
% caption: The subgroup criterion as a decision procedure; the single test
% $xy^{-1}\in H$ folds closure under products and closure under inverses into one.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=34mm, minimum height=9mm, align=center},
  dec/.style={draw, diamond, aspect=2.2, inner sep=1pt, align=center},
  term/.style={draw, minimum width=26mm, minimum height=9mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[box] (a) at (0,0) {$H$ nonempty};
  \node[dec] (b) at (0,-1.9) {closed under\\the test map};
  \node[term, draw=acc, text=acc] (yes) at (4.4,-1.9) {$H$ a subgroup};
  \node[term] (no) at (0,-4.0) {not a subgroup};
  \draw[->, acc, thick] (a) -- (b);
  \draw[->, acc, thick] (b) -- (yes) node[midway, above, font=\scriptsize]{yes};
  \draw[->, black, thick] (b) -- (no) node[midway, right, font=\scriptsize]{no};
  \node[align=left, font=\scriptsize, anchor=west] at (-3.6,-1.9)
    {test each\\pair $x$, $y$};
\end{tikzpicture}
$$

The finite case is what makes checking small groups mechanical: for a subset of
a finite group, only the multiplication table needs to stay inside the subset.
Two more inheritance facts round out the basic theory: the intersection of any
nonempty family of subgroups is again a subgroup (closure passes to the
intersection), while the union of two subgroups is a subgroup only in the
degenerate case where one contains the other.

## Subgroups attached to a subset

Every nonempty subset $A \subseteq G$ generates three subgroups measuring how
$G$ commutes with $A$. Fix $A$ throughout.

> **Definition (Centralizer).** The **centralizer** of $A$ in $G$ is
> $C_G(A) = \{ g \in G \mid gag^{-1} = a \text{ for all } a \in A \}$, the set
> of elements commuting with every element of $A$ (since $gag^{-1} = a$ is
> equivalent to $ga = ag$).

> **Definition (Center).** The **center** of $G$ is
> $Z(G) = \{ g \in G \mid gx = xg \text{ for all } x \in G \}$, the elements
> commuting with everything. Note $Z(G) = C_G(G)$.

> **Definition (Normalizer).** Writing $gAg^{-1} = \{ gag^{-1} \mid a \in A \}$,
> the **normalizer** of $A$ in $G$ is
> $N_G(A) = \{ g \in G \mid gAg^{-1} = A \}$, the elements that carry the set
> $A$ to itself under conjugation.

Each of these is a subgroup. The verification for $C_G(A)$ is the template.[^df-cent]
The identity commutes with everything, so $1 \in C_G(A)$ and the set is
nonempty. If $x, y \in C_G(A)$, then from $yay^{-1} = a$ one recovers
$y^{-1}ay = a$, so $y^{-1} \in C_G(A)$; and

$$
(xy)a(xy)^{-1} = x(yay^{-1})x^{-1} = xax^{-1} = a,
$$

so $xy \in C_G(A)$. The same steps, with conjugation of the whole set $A$ in
place of a single element, prove $N_G(A) \leq G$; and $Z(G) = C_G(G)$ is the
centralizer of the largest possible subset.

These three sit in a fixed chain. If $g$ centralizes $A$ then $gag^{-1} = a \in A$
for every $a$, so $g$ certainly sends $A$ to $A$: $C_G(A) \leq N_G(A)$. And
$Z(G)$ commutes with everything, in particular with each element of $A$, so
$Z(G) \leq C_G(A)$.

> **Invariant.** For any nonempty $A \subseteq G$,
> $$
> Z(G) \;\leq\; C_G(A) \;\leq\; N_G(A) \;\leq\; G.
> $$

$$
% caption: The canonical chain of subgroups attached to a subset $A$; each is
% contained in the next, with the center innermost and $G$ outermost.
\begin{tikzpicture}[font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=acc!5, draw=black] (-3.6,-2.3) rectangle (3.6,2.3);
  \node[anchor=north east, black] at (3.5,2.2) {$G$};
  \draw[fill=acc!9, draw=black] (-2.7,-1.75) rectangle (2.7,1.75);
  \node[anchor=north east, black] at (2.6,1.65) {$N_G(A)$};
  \draw[fill=acc!13, draw=black] (-1.8,-1.2) rectangle (1.8,1.2);
  \node[anchor=north east, black] at (1.7,1.1) {$C_G(A)$};
  \draw[fill=acc!18, draw=acc] (-0.95,-0.62) rectangle (0.95,0.62);
  \node[text=acc] at (0,0) {$Z(G)$};
\end{tikzpicture}
$$

> **Worked example.** The centralizer, normalizer, and center of the rotation
> subgroup of $D_8$. Let $D_8 = \langle r, s \rangle$ with $r^4 = s^2 = 1$ and
> $srs = r^{-1}$, and let $A = \{1, r, r^2, r^3\}$. All powers of $r$ commute, so
> $A \leq C_{D_8}(A)$. The reflection $s$ satisfies $sr = r^{-1}s \neq rs$, so
> $s \notin C_{D_8}(A)$; the same holds for every element $sr^i$. Hence
> $C_{D_8}(A) = A$. For the normalizer, $sAs^{-1} = \{1, r^3, r^2, r\} = A$, so
> $s \in N_{D_8}(A)$; since $r$ and $s$ generate $D_8$ and both normalize $A$,
> every element does, giving $N_{D_8}(A) = D_8$. The center is smaller still:
> $Z(D_8) \leq C_{D_8}(A) = A$, and among the rotations only $1$ and $r^2$
> commute with $s$, so $Z(D_8) = \{1, r^2\}$.[^df-d8]

$$
% caption: For $A$ the rotation subgroup of $D_8$, the chain collapses to
% $Z(D_8) = \{1, r^2\}$ inside $C_{D_8}(A) = A$ inside $N_{D_8}(A) = D_8$.
\begin{tikzpicture}[font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=acc!5, draw=black] (-3.5,-1.6) rectangle (3.5,1.6);
  \node[anchor=north west, black] at (-3.4,1.5) {$N_{D_8}(A) = D_8$};
  \draw[fill=acc!12, draw=black] (-2.1,-1.05) rectangle (2.1,1.05);
  \node[anchor=north, black] at (0,0.95) {$C_{D_8}(A) = A$};
  \node[font=\scriptsize] at (-1.3,0.35) {$1$};
  \node[font=\scriptsize] at (-0.45,0.35) {$r$};
  \node[font=\scriptsize] at (0.45,0.35) {$r^2$};
  \node[font=\scriptsize] at (1.35,0.35) {$r^3$};
  \draw[fill=acc!20, draw=acc] (-1.0,-0.75) rectangle (1.0,-0.05);
  \node[text=acc, font=\scriptsize] at (0,-0.4) {$Z(D_8) = < r^2 >$};
  \node[font=\scriptsize, black] at (2.65,-0.9) {$s$, $sr$, $sr^2$, $sr^3$};
\end{tikzpicture}
$$

> **Worked example.** The center of $S_3$, cutting the cases by order. Let
> $A = \{1, (1\,2)\}$ in $S_3$. Since an element commutes with its own powers,
> $A \le C_{S_3}(A)$, so $2$ divides $|C_{S_3}(A)|$, which in turn divides
> $|S_3| = 6$ (this divisibility is Lagrange's theorem, proved in the
> [cosets lesson](/abstract-algebra/subgroups-and-quotients/cosets-lagrange-and-normal-subgroups);
> here it only prunes cases). The options are $2$ or $6$. If $6$, then $(1\,2)$
> would commute with everything, but $(1\,2)(1\,2\,3) = (2\,3)$ while
> $(1\,2\,3)(1\,2) = (1\,3)$. So $C_{S_3}(A) = A$. The normalizer is no bigger:
> $\sigma \in N_{S_3}(A)$ requires $\sigma (1\,2) \sigma^{-1} = (1\,2)$, which is
> again the centralizer condition, so $N_{S_3}(A) = A$ as well. And
> $Z(S_3) \le C_{S_3}(A) = A$ with $(1\,2) \notin Z(S_3)$ forces $Z(S_3) = 1$:
> the smallest non-abelian group has trivial center. Order constraints cut the
> cases, then one explicit conjugation settles what remains, with no need to
> check every element.[^df-s3]

## Stabilizers and kernels of actions

The same subgroups appear from the direction of
[group actions](/abstract-algebra/groups-and-symmetry/homomorphisms-and-group-actions).
When $G$ acts on a set $S$ and $s \in S$ is fixed, two subsets of $G$ record how
$G$ moves $s$ and how $G$ moves all of $S$.

> **Definition (Stabilizer).** For $G$ acting on $S$ and $s \in S$, the
> **stabilizer** of $s$ is $G_s = \{ g \in G \mid g \cdot s = s \}$.

> **Definition (Kernel of an action).** The **kernel** of the action is
> $\{ g \in G \mid g \cdot s = s \text{ for all } s \in S \}$, the elements
> acting trivially on every point.

Both are subgroups, by an argument that mirrors the centralizer proof with the
action axioms replacing associativity.[^df-stab] For the stabilizer:
$1 \cdot s = s$ so $1 \in G_s$; if $y \in G_s$ then
$s = 1 \cdot s = (y^{-1}y) \cdot s = y^{-1} \cdot (y \cdot s) = y^{-1} \cdot s$,
so $y^{-1} \in G_s$; and if $x, y \in G_s$ then
$(xy) \cdot s = x \cdot (y \cdot s) = x \cdot s = s$. The kernel is the
intersection $\bigcap_{s \in S} G_s$, so it is a subgroup too.

$$
% caption: $D_8$ acting on the four vertices of a square; the stabilizer of a
% vertex is the order-two subgroup generated by the reflection through it.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \coordinate (v1) at (-1.4,1.4);
  \coordinate (v2) at (1.4,1.4);
  \coordinate (v3) at (1.4,-1.4);
  \coordinate (v4) at (-1.4,-1.4);
  \draw[black, thick] (v1) -- (v2) -- (v3) -- (v4) -- cycle;
  % reflection axis through vertex v1 and v3 (the main diagonal)
  \draw[acc, dashed, thick] (-1.95,1.95) -- (1.95,-1.95);
  \node[acc, anchor=west, font=\scriptsize] at (2.1,-1.75) {axis of $t$};
  \foreach \p/\lab/\pos in {v1/1/left, v2/2/above right, v3/3/below, v4/4/below left}
    \fill (\p) circle (2.4pt) node[\pos, font=\scriptsize] {\lab};
  \node[draw=acc, text=acc, align=center, font=\scriptsize, anchor=west]
    at (3.0,0) {$G_1 = < t >$\\stabilizer\\of vertex $1$};
  \draw[->, acc] (2.0,0) -- (2.95,0);
\end{tikzpicture}
$$

For the square, $D_8$ permutes the four vertices. The stabilizer of a vertex is
$\{1, t\}$ where $t$ is the reflection through that vertex and the center; the
kernel of this action is trivial, since only the identity fixes all four
vertices. A second action of $D_8$, on the two diagonals $\{\{1,3\}, \{2,4\}\}$,
has kernel $\{1, s, r^2, sr^2\}$, and this larger kernel is why that action
fails to be faithful.

The three subgroups from the previous section are special cases. Let $G$ act on
its own power set $\mathcal{P}(G)$ by conjugation, $g : B \mapsto gBg^{-1}$. Then
$N_G(A)$ is precisely the stabilizer of the point $A \in \mathcal{P}(G)$; letting
$N_G(A)$ act on $A$ by conjugation, $C_G(A)$ is the kernel; and $Z(G)$ is the
kernel of $G$ acting on itself by conjugation.[^df-unify]

| Subgroup | Elements | As a stabilizer/kernel |
| --- | --- | --- |
| $C_G(A)$ | commute with each $a \in A$ | kernel of $N_G(A)$ acting on $A$ by conjugation |
| $Z(G)$ | commute with all of $G$ | kernel of $G$ acting on $G$ by conjugation |
| $N_G(A)$ | send $A$ to $A$ | stabilizer of $A \in \mathcal{P}(G)$ under conjugation |
| $G_s$ | fix the point $s$ | stabilizer of $s$ in the action on $S$ |

## Interpreting the center and normalizer

The center measures how far $G$ is from abelian: $Z(G) = G$ exactly when $G$ is
abelian, and $Z(D_8) = \{1, r^2\}$ says $D_8$ is far from it. The normalizer
measures how close a subgroup is to being **normal**, the property that makes
quotient groups possible, developed in
[cosets and normal subgroups](/abstract-algebra/subgroups-and-quotients/cosets-lagrange-and-normal-subgroups).
The stabilizer, paired with the orbit, gives the counting law behind the
[Sylow theorems](/abstract-algebra/group-actions-and-sylow/sylow-theorems).

[^df-def]: **Dummit & Foote**, _Abstract Algebra_, §2.1 — Definition and Examples: the definition of a subgroup, the notation $H \leq G$, and the cancellation argument showing the identity and inverses of $H$ agree with those of $G$.
[^df-crit]: **Dummit & Foote**, §2.1, Proposition 1 — the Subgroup Criterion, including the finite-group refinement to closure under multiplication.
[^df-cent]: **Dummit & Foote**, §2.2 — Centralizers and Normalizers: definitions of $C_G(A)$, $Z(G)$, $N_G(A)$ and the proofs that each is a subgroup, with $C_G(A) \leq N_G(A)$.
[^df-d8]: **Dummit & Foote**, §2.2, Examples 2–4 — the computation of $C_{D_8}(A)$, $N_{D_8}(A)$, and $Z(D_8)$ for $A$ the rotation subgroup.
[^df-s3]: **Dummit & Foote**, §2.2, Example 5 — $C_{S_3}(\{1, (1\,2)\}) = N_{S_3}(\{1, (1\,2)\}) = \{1, (1\,2)\}$ and $Z(S_3) = 1$, using Lagrange to cut the cases.
[^df-stab]: **Dummit & Foote**, §2.2 — Stabilizers and Kernels of Group Actions: the stabilizer $G_s$ and the kernel of an action are subgroups, proved by the action axioms.
[^df-unify]: **Dummit & Foote**, §2.2 — the unification of $N_G(A)$, $C_G(A)$, and $Z(G)$ as stabilizers and kernels of conjugation actions.