Subgroups and Their Substructures

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

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.¹

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:

$Z \leq Q \leq R$ under addition; the even integers $2 Z \leq Z$ .
In the dihedral group $D_{2 n}$ , 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 $Z^{+}$ are not a subgroup of $Z$ : they miss the identity $0$ and are not closed under inverses.
$Q ∖ {0}$ under multiplication is not a subgroup of $R$ 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.

The proof is short.² Take any $x \in H$ (available since $H$ is nonempty). Applying the condition to $y = x$ gives $x x^{- 1} = 1 \in H$ . 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 $x (y^{- 1})^{- 1} = x y \in H$ 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 $x^{- 1} = x^{b - a - 1} \in H$ .

The subgroup criterion as a decision procedure; the single test

x y^{- 1} \in H

folds closure under products and closure under inverses into one.

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.

Each of these is a subgroup. The verification for $C_{G} (A)$ is the template.³ 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 $y a y^{- 1} = a$ one recovers $y^{- 1} a y = a$ , so $y^{- 1} \in C_{G} (A)$ ; and

(x y) a (x y)^{- 1} = x (y a y^{- 1}) x^{- 1} = x a x^{- 1} = a,

so $x y \in C_{G} (A)$ . 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 $g a g^{- 1} = a \in A$ 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 $Z (G) \leq C_{G} (A)$ .

The canonical chain of subgroups attached to a subset

A

; each is contained in the next, with the center innermost and

G

outermost.

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}

Worked example. The center of $S_{3}$ , cutting the cases by order. Let $A = {1, (12)}$ in $S_{3}$ . Since an element commutes with its own powers, $A \leq C_{S_{3}} (A)$ , 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; here it only prunes cases). The options are $2$ or $6$ . If $6$ , then $(12)$ would commute with everything, but $(12) (123) = (23)$ while $(123) (12) = (13)$ . So $C_{S_{3}} (A) = A$ . The normalizer is no bigger: $σ \in N_{S_{3}} (A)$ requires $σ (12) σ^{- 1} = (12)$ , which is again the centralizer condition, so $N_{S_{3}} (A) = A$ as well. And $Z (S_{3}) \leq C_{S_{3}} (A) = A$ with $(12) \in / Z (S_{3})$ 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.⁵

Stabilizers and kernels of actions

The same subgroups appear from the direction of 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 $S$ .

Both are subgroups, by an argument that mirrors the centralizer proof with the action axioms replacing associativity.⁶ 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 $(x y) \cdot s = x \cdot (y \cdot s) = x \cdot s = s$ . The kernel is the intersection $⋂_{s \in S} G_{s}$ , so it is a subgroup too.

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.

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 ${1, s, r^{2}, s r^{2}}$ , 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 $P (G)$ by conjugation, $g : B \mapsto g B g^{- 1}$ . Then $N_{G} (A)$ is precisely the stabilizer of the point $A \in P (G)$ ; 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.⁷

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 $A \in P (G)$ 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. The stabilizer, paired with the orbit, gives the counting law behind the Sylow theorems.

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 $G$ . ↩
Dummit & Foote, §2.1, Proposition 1 — the Subgroup Criterion, including the finite-group refinement to closure under multiplication. ↩
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 $C_{G} (A) \leq N_{G} (A)$ . ↩
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. ↩
Dummit & Foote, §2.2, Example 5 — $C_{S_{3}} ({1, (12)}) = N_{S_{3}} ({1, (12)}) = {1, (12)}$ and $Z (S_{3}) = 1$ , using Lagrange to cut the cases. ↩
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. ↩
Dummit & Foote, §2.2 — the unification of $N_{G} (A)$ , $C_{G} (A)$ , and $Z (G)$ as stabilizers and kernels of conjugation actions. ↩