Helpful context:

The central idea of abstract algebra is this: by stripping away everything except a short list of axioms, we can prove theorems that apply to integers, rotations, permutations, and symmetries all at once. One proof, infinite applications.

Here is a concrete example of what that buys. Lagrange’s theorem states that if $H$ is a subgroup of a finite group $G$, then $|H|$ divides $|G|$. This single fact, proved once from four axioms, immediately implies: the order of any element of $\mathbb{Z}_n$ divides $n$; Fermat’s little theorem ($a^{p-1} \equiv 1 \pmod{p}$ for prime $p$); the size of any symmetry subgroup of a crystal divides the total symmetry group of that crystal. Without abstraction, each would require its own separate proof. With it, all three are special cases of one theorem.

The axioms are minimal and chosen precisely so that the interesting structure emerges. The cost is loss of concreteness - a group element is not a number or a matrix, it is whatever satisfies the axioms, and proofs must use nothing beyond what the axioms say. The gain is universality: every theorem holds for all groups, everywhere the axioms hold.

What follows is a thorough development of group theory from first principles: the axioms, basic structural properties with their proofs, subgroups, cosets, Lagrange’s theorem, cyclic groups, isomorphisms, normal subgroups, and quotient groups. The subject is rich enough to fill several textbooks; the goal here is to develop each concept carefully, with proofs, so that the structure becomes visible rather than merely named.

The Group Axioms

A group is a pair $(G, \cdot)$ where $G$ is a set and $\cdot : G \times G \to G$ is a binary operation satisfying four axioms:

  1. Closure. For all $a, b \in G$, the product $a \cdot b \in G$.
  2. Associativity. For all $a, b, c \in G$, $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
  3. Identity. There exists an element $e \in G$ such that $e \cdot a = a \cdot e = a$ for all $a \in G$.
  4. Inverses. For every $a \in G$, there exists an element $a^{-1} \in G$ such that $a \cdot a^{-1} = a^{-1} \cdot a = e$.

A group is abelian (or commutative) if it additionally satisfies commutativity: $a \cdot b = b \cdot a$ for all $a, b \in G$. Non-abelian groups exist and are central to the theory; the symmetric groups are the canonical examples.

Closure is sometimes implicit - if the operation is defined as a function $G \times G \to G$, it maps into $G$ by definition. The axiom is stated explicitly to emphasize that the product of two elements must stay within the group, not escape to some larger set. This becomes substantive when checking whether a subset forms a subgroup.

Examples

$(\mathbb{Z}, +)$. The integers under addition form an abelian group. Closure: sum of integers is an integer. Associativity: addition is associative. Identity: $0$. Inverses: the inverse of $n$ is $-n$. This is abelian because $m + n = n + m$.

$(\mathbb{R} \setminus \{0\}, \times)$. The nonzero reals under multiplication. Identity: $1$. Inverse of $x$: $1/x$. Zero must be excluded because $0$ has no multiplicative inverse. This is abelian.

$(\mathbb{Z}_n, +_n)$. The integers modulo $n$ under addition modulo $n$, for $n \geq 1$. Elements: $\{0, 1, \ldots, n-1\}$. Identity: $0$. Inverse of $k$: $n - k$ (with $n - 0 = 0$). Abelian.

$S_n$, the symmetric group. Let $S_n$ be the set of all bijections $\sigma : \{1, \ldots, n\} \to \{1, \ldots, n\}$ under composition. These are permutations of $n$ elements. Identity: the identity permutation $\text{id}$. Inverse of $\sigma$: the inverse bijection $\sigma^{-1}$. Closure and associativity follow from properties of function composition. For $n \geq 3$, $S_n$ is non-abelian: the permutations $(12)$ and $(13)$ do not commute. The group $S_n$ has order $n!$.

$GL(n, \mathbb{R})$, the general linear group. The set of all invertible $n \times n$ real matrices under matrix multiplication. Identity: the identity matrix $I_n$. Inverse: the matrix inverse $A^{-1}$. Closure: the product of two invertible matrices is invertible (since $\det(AB) = \det(A)\det(B) \neq 0$ when both factors are nonzero). This group is non-abelian for $n \geq 2$.

Basic Properties

Uniqueness of the Identity

Theorem. The identity element of a group is unique.

Proof. Suppose $e$ and $e'$ are both identities. Since $e$ is an identity, $e \cdot e' = e'$. Since $e'$ is an identity, $e \cdot e' = e$. Therefore $e = e'$. $\square$

The proof is a one-line calculation. It uses both the left and right identity properties: $e \cdot e' = e'$ (because $e$ is an identity acting on $e'$) and $e \cdot e' = e$ (because $e'$ is an identity acting on $e$, reading right to left). Both hold by the identity axiom, and equating them gives uniqueness.

Uniqueness of Inverses

Theorem. For each $a \in G$, the inverse $a^{-1}$ is unique.

Proof. Suppose $b$ and $c$ are both inverses of $a$, so $a \cdot b = b \cdot a = e$ and $a \cdot c = c \cdot a = e$. Then:

$$b = b \cdot e = b \cdot (a \cdot c) = (b \cdot a) \cdot c = e \cdot c = c.$$

The steps use: identity, the fact that $c$ is an inverse of $a$, associativity, the fact that $b$ is an inverse of $a$, and identity again. $\square$

Cancellation Law

Theorem. If $a \cdot b = a \cdot c$, then $b = c$. Likewise, if $b \cdot a = c \cdot a$, then $b = c$.

Proof (left cancellation). Multiply both sides on the left by $a^{-1}$:

$$a^{-1} \cdot (a \cdot b) = a^{-1} \cdot (a \cdot c).$$

By associativity, $(a^{-1} \cdot a) \cdot b = (a^{-1} \cdot a) \cdot c$, so $e \cdot b = e \cdot c$, giving $b = c$. $\square$

The cancellation law looks obvious from experience with numbers, but it fails for some algebraic structures (like integers under multiplication: $0 \cdot 2 = 0 \cdot 3$ does not imply $2 = 3$). In a group, inverses always exist, which is what makes cancellation valid.

Inverse of a Product

Theorem. For all $a, b \in G$, $(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$.

Proof. It suffices to verify that $b^{-1} \cdot a^{-1}$ is an inverse of $a \cdot b$, i.e., that their product is $e$ on each side.

$$(a \cdot b) \cdot (b^{-1} \cdot a^{-1}) = a \cdot (b \cdot b^{-1}) \cdot a^{-1} = a \cdot e \cdot a^{-1} = a \cdot a^{-1} = e.$$

Similarly $(b^{-1} \cdot a^{-1}) \cdot (a \cdot b) = e$. By uniqueness of inverses, $(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$. $\square$

The reversal of order is the crucial point: dressing in reverse order. If you put on socks then shoes, you must remove shoes then socks.

Order of an Element

The order of an element $a \in G$, written $\text{ord}(a)$, is the smallest positive integer $k$ such that $a^k = e$, where $a^k$ means $a$ applied $k$ times (i.e., $a \cdot a \cdots a$, $k$ factors). If no such $k$ exists, $a$ has infinite order.

The order of a group $|G|$ is the cardinality of the set $G$. A group is finite if $|G| < \infty$.

For example, in $\mathbb{Z}_6$ under addition mod $6$: $\text{ord}(2) = 3$ because $2 + 2 + 2 = 6 \equiv 0$. And $\text{ord}(1) = 6$ because the smallest $k$ with $k \cdot 1 \equiv 0 \pmod{6}$ is $k = 6$.

Subgroups

A subset $H \subseteq G$ is a subgroup of $G$, written $H \leq G$, if $H$ is itself a group under the same operation inherited from $G$.

The Subgroup Test. $H \leq G$ if and only if $H$ is nonempty and for all $a, b \in H$, $a \cdot b^{-1} \in H$.

Proof. If $H \leq G$, then $H$ is nonempty (it contains the identity), and closure plus inverses give $a \cdot b^{-1} \in H$.

Conversely, suppose $H$ is nonempty and closed under $a \mapsto a \cdot b^{-1}$. Pick any $h \in H$. Setting $a = b = h$ gives $h \cdot h^{-1} = e \in H$, so the identity is in $H$. For any $h \in H$, setting $a = e$ gives $e \cdot h^{-1} = h^{-1} \in H$, so $H$ is closed under inverses. For closure under multiplication: given $a, b \in H$, we have $b^{-1} \in H$, and then $a \cdot (b^{-1})^{-1} = a \cdot b \in H$. Associativity is inherited from $G$. $\square$

The subgroup test compresses all four axioms into a single condition that is efficient to verify.

Key Examples of Subgroups

Trivial subgroup. $\{e\} \leq G$ for any group $G$. It contains only the identity.

The center $Z(G)$. Define $Z(G) = \{a \in G \mid ax = xa \text{ for all } x \in G\}$. This is the set of elements that commute with everything.

Proof that $Z(G) \leq G$. $Z(G)$ is nonempty since $e \in Z(G)$ (as $ex = x = xe$). Let $a, b \in Z(G)$. Then $ab^{-1}$ commutes with any $x$: we need $(ab^{-1})x = x(ab^{-1})$. Since $b \in Z(G)$, we have $bx = xb$, so $b^{-1}x = xb^{-1}$ (multiply on both sides by $b^{-1}$). Since $a \in Z(G)$, $ax = xa$. Then $(ab^{-1})x = a(b^{-1}x) = a(xb^{-1}) = (ax)b^{-1} = (xa)b^{-1} = x(ab^{-1})$. So $ab^{-1} \in Z(G)$. By the subgroup test, $Z(G) \leq G$. $\square$

The center is always a normal subgroup (we will see why later), and $G$ is abelian if and only if $Z(G) = G$.

The centralizer $C_G(a)$. For a fixed $a \in G$, define $C_G(a) = \{g \in G \mid ga = ag\}$. By an argument similar to the center, $C_G(a) \leq G$. Note $Z(G) = \bigcap_{a \in G} C_G(a)$.

Intersection of subgroups. If $\{H_i\}$ is any collection of subgroups of $G$, then $\bigcap_i H_i \leq G$.

Proof. The intersection is nonempty since $e \in H_i$ for all $i$. If $a, b \in \bigcap H_i$, then $a, b \in H_i$ for each $i$, so $ab^{-1} \in H_i$ for each $i$ (since each $H_i$ is a subgroup), hence $ab^{-1} \in \bigcap H_i$. $\square$

Cosets and Lagrange’s Theorem

Let $H \leq G$ and $a \in G$. The left coset of $H$ with representative $a$ is:

$$aH = \{ah \mid h \in H\}.$$

Similarly the right coset is $Ha = \{ha \mid h \in H\}$. In an abelian group these coincide, but in general they need not.

Cosets Partition G

Theorem. The left cosets of $H$ in $G$ partition $G$: every element of $G$ belongs to exactly one left coset.

Proof. First, every $g \in G$ belongs to some coset: $g = g \cdot e \in gH$ (since $e \in H$).

Second, any two left cosets are either equal or disjoint. Suppose $aH \cap bH \neq \emptyset$. Pick $x \in aH \cap bH$. Then $x = ah_1 = bh_2$ for some $h_1, h_2 \in H$, giving $a = bh_2 h_1^{-1}$. Since $H$ is a subgroup, $h_2 h_1^{-1} \in H$, so $a \in bH$. For any $ah \in aH$, we have $ah = b(h_2 h_1^{-1} h) \in bH$ since $h_2 h_1^{-1} h \in H$. So $aH \subseteq bH$. By symmetry $bH \subseteq aH$, hence $aH = bH$. $\square$

All Cosets Have the Same Size

Lemma. For any $a \in G$, $|aH| = |H|$.

Proof. The map $\phi: H \to aH$ defined by $\phi(h) = ah$ is a bijection. It is surjective by definition of $aH$. It is injective: if $ah = ah'$ then $h = h'$ by left cancellation. $\square$

Lagrange’s Theorem

Theorem (Lagrange). If $G$ is a finite group and $H \leq G$, then $|H|$ divides $|G|$.

Proof. The left cosets of $H$ partition $G$ (proved above) and each coset has size $|H|$ (proved above). Let $k$ be the number of distinct left cosets. Since the cosets are pairwise disjoint and cover $G$, we have:

$$|G| = k \cdot |H|.$$

Therefore $|H|$ divides $|G|$. $\square$

The number $k = |G|/|H|$ is called the index of $H$ in $G$, written $[G : H]$.

Consequences of Lagrange’s Theorem

Corollary 1. The order of any element $a \in G$ divides $|G|$.

Proof. The cyclic subgroup $\langle a \rangle = \{e, a, a^2, \ldots, a^{k-1}\}$ where $k = \text{ord}(a)$ is a subgroup of $G$ of order $k$. By Lagrange, $k$ divides $|G|$. $\square$

Corollary 2. For any finite group $G$ and any $a \in G$, $a^{|G|} = e$.

Proof. Write $|G| = k \cdot \text{ord}(a)$ for some integer $k$. Then $a^{|G|} = (a^{\text{ord}(a)})^k = e^k = e$. $\square$

Corollary 3 (Fermat’s Little Theorem). If $p$ is prime and $p \nmid a$, then $a^{p-1} \equiv 1 \pmod{p}$.

Proof. The group $(\mathbb{Z}_p \setminus \{0\}, \times)$ is a group of order $p - 1$ (it is the group of units modulo a prime). By Corollary 2, $a^{p-1} \equiv 1 \pmod{p}$. $\square$

Corollary 4. A group of prime order $p$ is cyclic.

Proof. Let $|G| = p$ and pick any $a \neq e$ in $G$. The cyclic subgroup $\langle a \rangle$ has order $\text{ord}(a) > 1$ (since $a \neq e$), and by Lagrange, $\text{ord}(a)$ divides $p$. Since $p$ is prime, $\text{ord}(a) = p$, so $\langle a \rangle$ has order $p = |G|$, giving $G = \langle a \rangle$. $\square$

Cyclic Groups

A group $G$ is cyclic if there exists an element $a \in G$ such that $G = \langle a \rangle = \{a^k \mid k \in \mathbb{Z}\}$. The element $a$ is called a generator of $G$.

Every cyclic group is abelian: $a^m \cdot a^n = a^{m+n} = a^n \cdot a^m$.

The prototypical finite cyclic group is $\mathbb{Z}_n = \langle 1 \rangle$ under addition mod $n$ (here $a^k$ means $k \cdot 1 = k$ in additive notation). The prototypical infinite cyclic group is $\mathbb{Z} = \langle 1 \rangle$ under addition.

Subgroups of $\mathbb{Z}_n$

Theorem. The subgroups of $\mathbb{Z}_n$ are exactly the sets $\langle d \rangle = \{0, d, 2d, \ldots, n-d\}$ for each divisor $d$ of $n$.

Proof sketch. Let $H \leq \mathbb{Z}_n$. If $H = \{0\}$, then $H = \langle n \rangle$ (since $n \equiv 0$). Otherwise, let $d$ be the smallest positive element of $H$. We claim $H = \langle d \rangle$.

Since $H$ is a subgroup and $d \in H$, all multiples $kd \in H$, so $\langle d \rangle \subseteq H$.

For the reverse inclusion, take any $m \in H$. By division with remainder, $m = qd + r$ with $0 \leq r < d$. Since $qd \in H$ and $m \in H$, we have $r = m - qd \in H$. But $d$ was the smallest positive element of $H$, and $0 \leq r < d$, so $r = 0$. Thus $m = qd \in \langle d \rangle$, giving $H \subseteq \langle d \rangle$.

So $H = \langle d \rangle$ for some positive $d \in H$. For $H$ to be a subgroup of $\mathbb{Z}_n$, we need the order of $d$ in $\mathbb{Z}_n$ (which is $n/\gcd(d,n)$) to divide $n$, which requires $d \mid n$. $\square$

In particular, the number of subgroups of $\mathbb{Z}_n$ equals the number of divisors of $n$.

Generators of Cyclic Groups

Theorem. A cyclic group of order $n$ has exactly $\varphi(n)$ generators, where $\varphi$ is Euler’s totient function.

Proof sketch. In $\mathbb{Z}_n = \langle 1 \rangle$, the element $k$ generates $\mathbb{Z}_n$ if and only if $\langle k \rangle = \mathbb{Z}_n$, i.e., $\text{ord}(k) = n$. The order of $k$ in $\mathbb{Z}_n$ is $n / \gcd(k, n)$, which equals $n$ if and only if $\gcd(k, n) = 1$. The number of such $k$ in $\{1, \ldots, n\}$ is $\varphi(n)$ by definition. $\square$

Every Subgroup of a Cyclic Group is Cyclic

Theorem. If $G$ is cyclic, then every subgroup $H \leq G$ is cyclic.

Proof sketch. Write $G = \langle a \rangle$. Every element of $H$ has the form $a^m$ for some $m \in \mathbb{Z}$. Let $d$ be the smallest positive integer such that $a^d \in H$ (such $d$ exists if $H \neq \{e\}$ by well-ordering). Then $\langle a^d \rangle \subseteq H$. For any $a^m \in H$, write $m = qd + r$ with $0 \leq r < d$. Then $a^r = a^m \cdot (a^d)^{-q} \in H$, and by minimality of $d$, $r = 0$. So $a^m = (a^d)^q \in \langle a^d \rangle$. Thus $H = \langle a^d \rangle$. $\square$

Group Isomorphism

Two groups $G$ and $H$ are isomorphic, written $G \cong H$, if there exists a bijection $\varphi: G \to H$ satisfying:

$$\varphi(a \cdot b) = \varphi(a) \cdot \varphi(b) \quad \text{for all } a, b \in G.$$

Such a map $\varphi$ is called an isomorphism. It preserves the group operation: multiplying then mapping gives the same result as mapping then multiplying. Isomorphic groups are structurally identical - they differ only in the names of their elements.

Group Homomorphisms

A map $\varphi: G \to H$ satisfying $\varphi(ab) = \varphi(a)\varphi(b)$ (without requiring bijectivity) is a homomorphism. Homomorphisms automatically preserve identity and inverses: $\varphi(e_G) = e_H$ and $\varphi(a^{-1}) = \varphi(a)^{-1}$.

Proof. $\varphi(e_G) = \varphi(e_G \cdot e_G) = \varphi(e_G) \cdot \varphi(e_G)$. Multiplying both sides on the right by $\varphi(e_G)^{-1}$ gives $\varphi(e_G) = e_H$. For inverses: $\varphi(a) \cdot \varphi(a^{-1}) = \varphi(a \cdot a^{-1}) = \varphi(e_G) = e_H$, so $\varphi(a^{-1}) = \varphi(a)^{-1}$. $\square$

Kernel and Image

The kernel of $\varphi: G \to H$ is $\ker(\varphi) = \{g \in G \mid \varphi(g) = e_H\}$.

The image of $\varphi$ is $\text{im}(\varphi) = \{\varphi(g) \mid g \in G\} \subseteq H$.

Theorem. $\ker(\varphi) \leq G$ and $\text{im}(\varphi) \leq H$.

Proof. For the kernel: $e_G \in \ker(\varphi)$ (since $\varphi(e_G) = e_H$), so it is nonempty. If $a, b \in \ker(\varphi)$, then $\varphi(ab^{-1}) = \varphi(a)\varphi(b^{-1}) = e_H \cdot \varphi(b)^{-1} = e_H^{-1} = e_H$, so $ab^{-1} \in \ker(\varphi)$. For the image: similar check using $\varphi(a)\varphi(b)^{-1} = \varphi(a)\varphi(b^{-1}) = \varphi(ab^{-1}) \in \text{im}(\varphi)$. $\square$

Moreover, $\ker(\varphi)$ is not merely a subgroup - it is a normal subgroup (defined in the next section): for any $g \in G$ and $k \in \ker(\varphi)$, $\varphi(gkg^{-1}) = \varphi(g)\varphi(k)\varphi(g)^{-1} = \varphi(g) \cdot e_H \cdot \varphi(g)^{-1} = e_H$, so $gkg^{-1} \in \ker(\varphi)$.

Isomorphisms Preserve Structure

Isomorphic groups have:

  • The same order: $|G| = |H|$.
  • The same element orders: $\text{ord}(a) = \text{ord}(\varphi(a))$ for all $a$.
  • The same number of elements of each order.
  • One is abelian if and only if the other is.

These are invariants of the isomorphism class. To show two groups are not isomorphic, it suffices to find one invariant that differs.

Classification by Isomorphism Type

Theorem. All cyclic groups of the same finite order are isomorphic: if $G = \langle a \rangle$ and $H = \langle b \rangle$ with $|G| = |H| = n$, then $G \cong H$ via $\varphi(a^k) = b^k$.

Corollary. There is, up to isomorphism, exactly one group of prime order $p$, namely $\mathbb{Z}_p$.

The First Isomorphism Theorem

Theorem (First Isomorphism Theorem). Let $\varphi: G \to H$ be a group homomorphism. Then:

$$G / \ker(\varphi) \cong \text{im}(\varphi).$$

The quotient group $G/\ker(\varphi)$ (defined in the next section) is isomorphic to the image. This theorem is one of the most used tools in group theory - it provides a canonical isomorphism whenever a homomorphism is given, and it says that every quotient group arises as an image of a homomorphism.

Automorphisms

An automorphism of a group $G$ is an isomorphism from $G$ to itself - a bijection $\phi: G \to G$ that preserves the group operation: $\phi(ab) = \phi(a)\phi(b)$ for all $a, b \in G$.

The set of all automorphisms of $G$, denoted $\text{Aut}(G)$, is itself a group under composition. Composition of two automorphisms is an automorphism (structure-preserving maps compose to structure-preserving maps), the identity map is an automorphism, and every automorphism has an inverse (which is also an automorphism).

Inner automorphisms. For any fixed $g \in G$, the map $\phi_g: x \mapsto gxg^{-1}$ is an automorphism called conjugation by $g$. It is an automorphism because $\phi_g(xy) = g(xy)g^{-1} = (gxg^{-1})(gyg^{-1}) = \phi_g(x)\phi_g(y)$. The set of all inner automorphisms $\text{Inn}(G) = \{\phi_g : g \in G\}$ forms a normal subgroup of $\text{Aut}(G)$.

The quotient $\text{Out}(G) = \text{Aut}(G) / \text{Inn}(G)$ is the outer automorphism group. An automorphism not of the form $x \mapsto gxg^{-1}$ is called an outer automorphism.

Examples.

  • $\text{Aut}(\mathbb{Z}_n) \cong \mathbb{Z}_n^*$: the automorphisms of the cyclic group $\mathbb{Z}_n$ are the maps $x \mapsto kx$ for $\gcd(k, n) = 1$. There are $\phi(n)$ such automorphisms where $\phi$ is Euler’s totient function.

  • $\text{Aut}(\mathbb{Z}) \cong \mathbb{Z}_2$: the only automorphisms of the integers are $x \mapsto x$ and $x \mapsto -x$.

  • For an abelian group, every inner automorphism is the identity (since $gxg^{-1} = x$ for all $x$). So $\text{Inn}(G)$ is trivial and $\text{Out}(G) \cong \text{Aut}(G)$.

The automorphism group of $S_n$. For $n \neq 6$, $\text{Aut}(S_n) \cong S_n$ (every automorphism is inner). For $n = 6$, there is a famous outer automorphism - one of the exceptional structures in group theory.

Characteristic subgroups. A subgroup $H \leq G$ is characteristic if $\phi(H) = H$ for every $\phi \in \text{Aut}(G)$. Characteristic subgroups are normal (since inner automorphisms are a special case), but not every normal subgroup is characteristic.

Normal Subgroups and Quotient Groups

A subgroup $N \leq G$ is normal in $G$, written $N \unlhd G$, if:

$$gNg^{-1} = N \quad \text{for all } g \in G.$$

Equivalently, $N \unlhd G$ if and only if $gN = Ng$ for all $g \in G$ - the left and right cosets of $N$ coincide. This is not the same as saying $gn = ng$ for all $g$ and $n$ (which would make $N \subseteq Z(G)$); it says $gng^{-1} \in N$ for all $n \in N$ and $g \in G$ (the conjugate of any element of $N$ is again in $N$).

Examples of normal subgroups:

  • $\{e\}$ and $G$ itself are always normal.
  • The center $Z(G) \unlhd G$ (since $g \cdot Z(G) \cdot g^{-1} = Z(G)$ by commutativity with all elements).
  • Every subgroup of an abelian group is normal.
  • The kernel of any homomorphism is normal (shown above).

Non-example. In $S_3$, the subgroup $H = \{e, (12)\}$ is not normal: $(123)(12)(123)^{-1} = (23) \notin H$.

Why Normality is Needed for Quotient Groups

Given a subgroup $N \leq G$, consider the set of left cosets $G/N = \{gN \mid g \in G\}$. We want to make this a group by defining $(gN)(hN) = (gh)N$. For this operation to be well-defined, we need the result to be independent of the choice of coset representatives. That is, if $gN = g’N$ and $hN = h’N$, we need $(gh)N = (g’h')N$.

If $g' = gn_1$ and $h' = hn_2$ for $n_1, n_2 \in N$, then $g’h' = gn_1 hn_2$. We need $gn_1 hn_2 N = ghN$, i.e., $(gh)^{-1}gn_1 hn_2 \in N$, i.e., $h^{-1}n_1 h \in N$. This holds for all $n_1 \in N$ and $h \in G$ precisely when $N$ is normal.

Theorem. If $N \unlhd G$, then $G/N$ is a group under the operation $(gN)(hN) = (gh)N$. The identity is $eN = N$, and the inverse of $gN$ is $g^{-1}N$.

The group $G/N$ is called the quotient group (or factor group) of $G$ by $N$. Its order is $|G/N| = [G:N] = |G|/|N|$.

Key Examples

$\mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$

The direct product $\mathbb{Z}_2 \times \mathbb{Z}_3 = \{(a, b) \mid a \in \mathbb{Z}_2, b \in \mathbb{Z}_3\}$ under componentwise addition has order $6$. It is generated by $(1, 1)$: the powers $(1,1), (0,2), (1,0), (0,1), (1,2), (0,0)$ exhaust all six elements, so $\mathbb{Z}_2 \times \mathbb{Z}_3$ is cyclic of order $6$.

Since $\mathbb{Z}_6$ is also cyclic of order $6$, the classification theorem gives $\mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$. This is the Chinese Remainder Theorem in group-theoretic form: $\gcd(2, 3) = 1$ implies $\mathbb{Z}_{mn} \cong \mathbb{Z}_m \times \mathbb{Z}_n$.

More generally, $\mathbb{Z}_{mn} \cong \mathbb{Z}_m \times \mathbb{Z}_n$ if and only if $\gcd(m, n) = 1$.

$S_3$ - The Smallest Non-Abelian Group

$S_3$ consists of the $6$ permutations of $\{1, 2, 3\}$: the identity $e$, three transpositions $(12), (13), (23)$, and two 3-cycles $(123), (132)$. It is non-abelian: $(12)(13) = (132)$ but $(13)(12) = (123)$.

$S_3$ has order $6$ and is not cyclic (no element has order $6$: the 3-cycles have order $3$, the transpositions have order $2$). So $S_3 \not\cong \mathbb{Z}_6$. In fact, up to isomorphism there are exactly two groups of order $6$: the cyclic group $\mathbb{Z}_6$ and the symmetric group $S_3$.

The normal subgroups of $S_3$ are $\{e\}$, $A_3 = \{e, (123), (132)\}$ (the alternating group, consisting of even permutations), and $S_3$ itself.

Cayley’s Theorem

Theorem (Cayley). Every group $G$ is isomorphic to a subgroup of the symmetric group $S_{|G|}$.

Proof sketch. For each $g \in G$, define $\lambda_g : G \to G$ by $\lambda_g(x) = gx$ (left multiplication by $g$). This is a bijection (with inverse $\lambda_{g^{-1}}$), so $\lambda_g \in S_{|G|}$ (viewing $G$ as a set of $|G|$ elements). The map $\Phi: G \to S_{|G|}$ defined by $\Phi(g) = \lambda_g$ is a group homomorphism: $\Phi(gh) = \lambda_{gh} = \lambda_g \circ \lambda_h = \Phi(g)\Phi(h)$. It is injective: if $\lambda_g = \lambda_h$, then $\lambda_g(e) = \lambda_h(e)$ gives $g = h$. Therefore $\Phi$ is an isomorphism from $G$ onto its image $\Phi(G) \leq S_{|G|}$. $\square$

Cayley’s theorem says that, in a precise sense, symmetric groups are universal: every abstract group appears inside some $S_n$. It justifies the intuition that groups are “really” groups of symmetries.

Summary

Concept Definition Key Result
Group Set with associative binary op, identity, inverses Foundation of abstract algebra
Abelian group Group with $ab = ba$ $(\mathbb{Z}, +)$, $(\mathbb{Z}_n, +_n)$
Order $\text{ord}(a)$ Smallest $k > 0$ with $a^k = e$ $\text{ord}(a)$ divides $
Subgroup $H \leq G$ Nonempty, closed under $ab^{-1}$ Center $Z(G)$, centralizer $C_G(a)$
Cosets $aH$ Translates of $H$ Partition $G$, all have size $
Lagrange’s theorem $ H
Cyclic group $\langle a \rangle$ Every element is a power of $a$ Has $\varphi(n)$ generators, subgroups cyclic
Homomorphism $\varphi$ $\varphi(ab) = \varphi(a)\varphi(b)$ $\ker(\varphi) \unlhd G$, $\text{im}(\varphi) \leq H$
Isomorphism $G \cong H$ Bijective homomorphism Preserves all group-theoretic structure
Normal subgroup $N \unlhd G$ $gNg^{-1} = N$ for all $g$ Needed for quotient group $G/N$
Quotient group $G/N$ Cosets with $(gN)(hN) = (gh)N$ Well-defined iff $N \unlhd G$
First isomorphism theorem $G/\ker(\varphi) \cong \text{im}(\varphi)$ Links quotients and images
Cayley’s theorem $G \hookrightarrow S_{ G

Why Four Axioms?

The group axioms are not arbitrary. They are the minimal set that makes the key theorems provable.

Drop closure, and you can’t compose elements and stay in the same structure - the object isn’t well-defined. Drop associativity, and the order of operations matters, making most algebraic manipulation impossible. Drop identity, and you lose the reference point from which inverses are defined. Drop inverses, and you have a monoid - a useful structure, but you lose the symmetry that makes groups powerful (you can’t “undo” operations).

Each axiom has a purpose. The fact that these four are both necessary and sufficient for a rich theory is one of the non-obvious miracles of mathematics. Weaker structures (monoids, semigroups) are useful but have weaker theorems. Stronger structures (abelian groups, fields) add more axioms and get more theorems. Groups sit at the exact point where the structure is rich enough to be interesting but minimal enough to be universal.

Recognizing groups in the wild. The test: find a set, find a binary operation, check the four axioms. If they hold, all of group theory applies for free. Symmetries of a triangle? Group (dihedral $D_3$). Invertible $n \times n$ matrices? Group ($GL_n$). Bijections from a set to itself? Group (symmetric group $S_n$). Strings under concatenation? Not a group (no inverses). Non-zero reals under multiplication? Group. Zero included? Not a group (0 has no multiplicative inverse).

The discipline of checking axioms, rather than relying on intuition, prevents errors that intuition hides.

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