Group Isomorphisms

When are two groups, described in completely different notations, secretly the same group? Consider a group of symmetries of a triangle. Compare it to a group of invertible residues modulo 7. These may look unrelated. Yet one question settles the matter completely: does a structure-preserving relabeling exist between them? This is the question of isomorphism. It is the precise sense in which abstract algebra allows us to say two objects "are the same." At the same time, it respects the fact that they may be built from entirely different underlying sets. The idea matters because group theory studies structure, not the incidental appearance of elements. Two groups can have different elements, different notations, even different cardinalities of description. Yet they can carry exactly the same multiplication pattern. Isomorphism is the formal tool that captures this sameness. It turns "same up to relabeling" into an honest equivalence relation on the (proper) class of all groups. Classifying groups up to isomorphism — rather than element-by-element — is the actual goal of most of group theory. This ranges from the classification of finite simple groups to the structure theorems for finitely generated abelian groups. The big picture: a Group Homomorphism is a map respecting the group operation. An isomorphism is a homomorphism that is also a bijection. So it has a two-sided inverse that is automatically also a homomorphism. This article develops the definition rigorously. It proves that isomorphism is an equivalence relation on any collection of groups. It also proves the fundamental fact that isomorphic groups share every group-theoretic property. These include order, element orders, abelian-ness, and subgroup lattice shape. We use these shared properties as invariants: quantities or properties that must agree if two groups are isomorphic. This gives a practical toolkit for proving groups are not isomorphic. It works without exhaustively checking every possible bijection. By the end, you will be able to state and prove that isomorphism is an equivalence relation. You will prove that isomorphisms preserve order, identity, inverses, abelian-ness, and cyclic structure. You will use invariants — order, element-order spectrum, abelian-ness, number of solutions to x2=ex^2=e — to separate groups efficiently. This groundwork is exactly what is needed for the Isomorphism Theorems. It also supports understanding Direct Products as a way of building new groups from old ones up to isomorphism.

Key takeaways
  • An isomorphism is a bijective Group Homomorphism; two groups are isomorphic (G≅HG \cong H) if one exists between them.
  • Isomorphism is an equivalence relation on any set of groups: reflexive, symmetric, transitive — proved rigorously below.
  • Every isomorphism automatically sends identity to identity, inverses to inverses, and preserves element order exactly.
  • Isomorphic groups agree on every "intrinsic" invariant: ∣G∣|G|, abelian-ness, cyclicity, the multiset of element orders, subgroup lattice shape.
  • Invariants let you disprove isomorphism quickly (e.g. different element-order spectra) without testing all bijections.
  • The Structure and Classification of Cyclic Groups shows every cyclic group of order nn is isomorphic to Z/nZ\mathbb{Z}/n\mathbb{Z}, the prototype application.
  • Key idea: isomorphism is exact structural sameness — everything expressible purely in terms of the group operation transfers perfectly across an isomorphism.

Where this lives today

Isomorphism classification is the organizing question behind the Classification of Finite Simple Groups, one of the largest collaborative theorems in mathematics, which lists every finite simple group up to isomorphism. In cryptography, the presumed computational difficulty of finding isomorphisms (or discrete logarithms, itself tied to cyclic group isomorphism with Z/nZ\mathbb{Z}/n\mathbb{Z}) underlies systems like Diffie–Hellman key exchange and elliptic-curve cryptography, where security relies on structural sameness being easy to state but hard to exploit computationally. In physics, isomorphic symmetry groups (e.g., SU(2)SU(2) and Spin(3)Spin(3) being a double cover rather than isomorphic, a subtlety that matters enormously) determine which physical theories share conserved quantities and selection rules. In computer science, the graph isomorphism problem is a close structural cousin, and group isomorphism testing for finite groups given by multiplication tables is itself a studied algorithmic problem with known sub-exponential algorithms. In chemistry, molecular symmetry groups are classified up to isomorphism to predict vibrational spectra and orbital hybridization patterns. Finally, every application of the Isomorphism Theorems — expressing quotient constructions as isomorphic to other explicit groups — depends on the basic equivalence-relation and invariant-preservation facts proved here.

Ask the tutor: Can you show step by step why U(8)U(8) is not isomorphic to Z/4Z\mathbb{Z}/4\mathbb{Z} using element orders?

DefinitionGroup Isomorphism

Let (G,⋅)(G, \cdot) and (H,∗)(H, *) be groups. A map φ:G→H\varphi: G \to H is a group isomorphism if: 1. φ\varphi is a homomorphism: φ(a⋅b)=φ(a)∗φ(b)\varphi(a \cdot b) = \varphi(a) * \varphi(b) for all a,b∈Ga, b \in G; 2. φ\varphi is a bijection (injective and surjective). If such a φ\varphi exists, we say GG and HH are isomorphic, written G≅HG \cong H. An isomorphism from GG to itself is called an automorphism of GG.

Intuition

Think of a group as a set of elements together with a "wiring diagram" recording how any two elements combine. An isomorphism is a relabeling of the elements — a dictionary translating names from GG into names in HH — under which the wiring diagrams match up perfectly. If you can find such a dictionary, then any true statement about GG phrased purely in terms of "⋅\cdot" translates, word for word, into a true statement about HH under "∗*". Conversely, if two groups differ in some property that must be shared by isomorphic groups (an invariant), no dictionary can exist. A useful mental picture: draw the Cayley table (multiplication table) of a small finite group. An isomorphism to another group of the same order is exactly a simultaneous relabeling of the rows, columns, and entries that turns one table into the other. This is why isomorphic finite groups literally have identical multiplication tables up to renaming — the "shape" of the table, not the names of the elements, is the invariant content of the group.

Figure
Two multiplication-table fragments matching under the relabeling 0↦e, 1↦r, 2↦r20\mapsto e,\,1\mapsto r,\,2\mapsto r^2
G = (Z/3Z, +) + 012 012 012 120 201 H = ⟨r⟩ ≤ D₃, r³ = e ∘ err² err² err² rr²e r²er φ(0)=e, φ(1)=r, φ(2)=r² — identical table shape
TheoremIsomorphism is an Equivalence Relation, and Isomorphisms Preserve Structure

Let G,H,KG, H, K be groups, all with identities denoted eG,eH,eKe_G, e_H, e_K respectively. Then:

(a) Equivalence relation. The relation "G≅HG \cong H" is reflexive, symmetric, and transitive on any collection of groups.

(b) Structure preservation. If φ:G→H\varphi: G \to H is an isomorphism, then: 1. φ(eG)=eH\varphi(e_G) = e_H; 2. φ(g−1)=φ(g)−1\varphi(g^{-1}) = \varphi(g)^{-1} for all g∈Gg \in G; 3. ord⁡(φ(g))=ord⁡(g)\operatorname{ord}(\varphi(g)) = \operatorname{ord}(g) for every g∈Gg \in G (including both being infinite); 4. GG is abelian if and only if HH is abelian; 5. GG is cyclic if and only if HH is cyclic, and generators correspond to generators; 6. φ\varphi restricts to a bijection between the subgroups of GG and the subgroups of HH, preserving containments and indices.

Proof

We give a complete proof, technique: direct verification from the definitions of homomorphism and bijection, using elementary set-theoretic and group-axiomatic reasoning with no forward appeal to later isomorphism-theorem machinery.

(a) Equivalence relation. Reflexive. The identity map id⁡G:G→G\operatorname{id}_G: G \to G satisfies id⁡G(ab)=ab=id⁡G(a)id⁡G(b)\operatorname{id}_G(ab) = ab = \operatorname{id}_G(a)\operatorname{id}_G(b), so it is a homomorphism, and it is trivially a bijection. Hence G≅GG \cong G. Symmetric. Suppose φ:G→H\varphi: G \to H is an isomorphism. Since φ\varphi is a bijection, it has a set-theoretic inverse φ−1:H→G\varphi^{-1}: H \to G, which is also a bijection. We must check φ−1\varphi^{-1} is a homomorphism. Let h1,h2∈Hh_1, h_2 \in H. Since φ\varphi is surjective, write h1=φ(g1)h_1 = \varphi(g_1), h2=φ(g2)h_2 = \varphi(g_2) for some g1,g2∈Gg_1, g_2 \in G. Then

φ−1(h1h2)=φ−1(φ(g1)φ(g2))=φ−1(φ(g1g2))=g1g2=φ−1(h1)φ−1(h2),\varphi^{-1}(h_1 h_2) = \varphi^{-1}(\varphi(g_1)\varphi(g_2)) = \varphi^{-1}(\varphi(g_1 g_2)) = g_1 g_2 = \varphi^{-1}(h_1)\varphi^{-1}(h_2),

where the second equality uses that φ\varphi is a homomorphism, and the third uses that φ−1∘φ=id⁡G\varphi^{-1}\circ \varphi = \operatorname{id}_G. Thus φ−1\varphi^{-1} is a homomorphism and a bijection, i.e. an isomorphism H→GH \to G. Hence H≅GH \cong G. Transitive. Suppose φ:G→H\varphi: G \to H and ψ:H→K\psi: H \to K are isomorphisms. The composite ψ∘φ:G→K\psi \circ \varphi : G \to K is a bijection, being a composite of bijections (composite of two injections is injective; composite of two surjections is surjective — standard set theory). It is a homomorphism because for a,b∈Ga,b \in G,

(ψ∘φ)(ab)=ψ(φ(ab))=ψ(φ(a)φ(b))=ψ(φ(a))ψ(φ(b))=(ψ∘φ)(a) (ψ∘φ)(b),(\psi\circ\varphi)(ab) = \psi(\varphi(ab)) = \psi(\varphi(a)\varphi(b)) = \psi(\varphi(a))\psi(\varphi(b)) = (\psi\circ\varphi)(a)\,(\psi\circ\varphi)(b),

using the homomorphism property of φ\varphi then of ψ\psi. Hence ψ∘φ\psi \circ \varphi is an isomorphism, so G≅KG \cong K. This establishes that ≅\cong is an equivalence relation on any collection of groups.

(b) Structure preservation. Fix an isomorphism φ:G→H\varphi: G \to H. (1) Identity. We have φ(eG)=φ(eGeG)=φ(eG)φ(eG)\varphi(e_G) = \varphi(e_G e_G) = \varphi(e_G)\varphi(e_G). Multiply both sides on the left by φ(eG)−1\varphi(e_G)^{-1} (which exists since HH is a group):

eH=φ(eG)−1φ(eG)=φ(eG)−1(φ(eG)φ(eG))=(φ(eG)−1φ(eG))φ(eG)=φ(eG).e_H = \varphi(e_G)^{-1}\varphi(e_G) = \varphi(e_G)^{-1}\big(\varphi(e_G)\varphi(e_G)\big) = \big(\varphi(e_G)^{-1}\varphi(e_G)\big)\varphi(e_G) = \varphi(e_G).

So φ(eG)=eH\varphi(e_G) = e_H. (This step only used that φ\varphi is a homomorphism, not bijectivity.) (2) Inverses. For g∈Gg \in G, using (1) and the homomorphism property,

φ(g)φ(g−1)=φ(gg−1)=φ(eG)=eH,φ(g−1)φ(g)=φ(g−1g)=φ(eG)=eH.\varphi(g)\varphi(g^{-1}) = \varphi(g g^{-1}) = \varphi(e_G) = e_H, \qquad \varphi(g^{-1})\varphi(g) = \varphi(g^{-1}g) = \varphi(e_G) = e_H.

By uniqueness of inverses in the group HH, φ(g−1)=φ(g)−1\varphi(g^{-1}) = \varphi(g)^{-1}. (3) Order preservation. First suppose gg has finite order n=ord⁡(g)n = \operatorname{ord}(g), meaning nn is the least positive integer with gn=eGg^n = e_G. By induction using the homomorphism property and (1), for any integer k≥1k \ge 1,

φ(gk)=φ(g)k\varphi(g^k) = \varphi(g)^k

(base case k=1k=1 trivial; inductive step φ(gk+1)=φ(gkg)=φ(gk)φ(g)=φ(g)kφ(g)=φ(g)k+1\varphi(g^{k+1}) = \varphi(g^k g) = \varphi(g^k)\varphi(g) = \varphi(g)^k \varphi(g) = \varphi(g)^{k+1}). Combined with (1) and (2) this extends to all integers kk, including negative and zero, by the standard convention g−k=(g−1)kg^{-k} = (g^{-1})^k and g0=eGg^0 = e_G. Hence φ(g)n=φ(gn)=φ(eG)=eH\varphi(g)^n = \varphi(g^n) = \varphi(e_G) = e_H, so ord⁡(φ(g))\operatorname{ord}(\varphi(g)) divides nn (order of an element divides any exponent giving the identity — a basic fact about cyclic subgroups, from Structure and Classification of Cyclic Groups); in particular ord⁡(φ(g))\operatorname{ord}(\varphi(g)) is finite, say m=ord⁡(φ(g))≤nm = \operatorname{ord}(\varphi(g)) \le n. Now run the same argument with φ−1\varphi^{-1} in place of φ\varphi (which is a homomorphism by part (a)'s symmetry proof): since φ(g)m=eH\varphi(g)^m = e_H, applying φ−1\varphi^{-1} gives φ−1(φ(g)m)=φ−1(eH)=eG\varphi^{-1}(\varphi(g)^m) = \varphi^{-1}(e_H) = e_G, and by the same induction φ−1(φ(g)m)=φ−1(φ(g))m=gm\varphi^{-1}(\varphi(g)^m) = \varphi^{-1}(\varphi(g))^m = g^m. So gm=eGg^m = e_G, forcing n∣mn \mid m, i.e. n≤mn \le m. Combining m≤nm \le n and n≤mn \le m gives m=nm = n: ord⁡(φ(g))=ord⁡(g)\operatorname{ord}(\varphi(g)) = \operatorname{ord}(g). If instead gg has infinite order, suppose toward contradiction that φ(g)\varphi(g) has finite order mm. Then φ(g)m=eH=φ(eG)\varphi(g)^m = e_H = \varphi(e_G), so φ(gm)=φ(eG)\varphi(g^m) = \varphi(e_G). Since φ\varphi is injective, gm=eGg^m = e_G, contradicting that gg has infinite order (as m≥1m \ge 1). Hence φ(g)\varphi(g) also has infinite order. (4) Abelian-ness. Suppose GG is abelian. For any h1,h2∈Hh_1, h_2 \in H, surjectivity gives g1,g2∈Gg_1, g_2 \in G with φ(gi)=hi\varphi(g_i) = h_i. Then

h1h2=φ(g1)φ(g2)=φ(g1g2)=φ(g2g1)=φ(g2)φ(g1)=h2h1,h_1 h_2 = \varphi(g_1)\varphi(g_2) = \varphi(g_1 g_2) = \varphi(g_2 g_1) = \varphi(g_2)\varphi(g_1) = h_2 h_1,

using commutativity of GG in the middle step. So HH is abelian. The converse follows by symmetry of ≅\cong (part (a)), applying the same argument to φ−1:H→G\varphi^{-1}: H \to G. (5) Cyclicity. Suppose G=⟨g0⟩G = \langle g_0 \rangle is cyclic, generated by g0g_0. We claim H=⟨φ(g0)⟩H = \langle \varphi(g_0)\rangle. Indeed, every h∈Hh \in H equals φ(g)\varphi(g) for some g∈Gg \in G (surjectivity), and since GG is cyclic, g=g0kg = g_0^k for some integer kk. Then, by the power law established in (3),

h=φ(g0k)=φ(g0)k∈⟨φ(g0)⟩.h = \varphi(g_0^k) = \varphi(g_0)^k \in \langle \varphi(g_0)\rangle.

So H⊆⟨φ(g0)⟩⊆HH \subseteq \langle \varphi(g_0)\rangle \subseteq H, giving equality: HH is cyclic, generated by φ(g0)\varphi(g_0). The converse again follows by applying this to φ−1\varphi^{-1}. (6) Subgroup correspondence. Let S≤GS \le G be a subgroup. We check φ(S)={φ(s):s∈S}\varphi(S) = \{\varphi(s): s \in S\} is a subgroup of HH: it is nonempty since eH=φ(eG)∈φ(S)e_H = \varphi(e_G) \in \varphi(S) (as eG∈Se_G \in S); it is closed under the operation since for s1,s2∈Ss_1, s_2 \in S, φ(s1)φ(s2)=φ(s1s2)∈φ(S)\varphi(s_1)\varphi(s_2) = \varphi(s_1 s_2) \in \varphi(S) because s1s2∈Ss_1 s_2 \in S (SS a subgroup); it is closed under inverses since φ(s)−1=φ(s−1)∈φ(S)\varphi(s)^{-1} = \varphi(s^{-1}) \in \varphi(S) by (2), as s−1∈Ss^{-1} \in S. So φ(S)≤H\varphi(S) \le H. Define Φ:{subgroups of G}→{subgroups of H}\Phi: \{\text{subgroups of } G\} \to \{\text{subgroups of } H\} by Φ(S)=φ(S)\Phi(S) = \varphi(S), and similarly Ψ(T)=φ−1(T)\Psi(T) = \varphi^{-1}(T) for subgroups T≤HT \le H (the same argument, applied to the isomorphism φ−1\varphi^{-1}, shows Ψ(T)≤G\Psi(T) \le G). We claim Ψ∘Φ=id⁡\Psi \circ \Phi = \operatorname{id} and Φ∘Ψ=id⁡\Phi \circ \Psi = \operatorname{id}, making Φ\Phi a bijection. Since φ\varphi is injective, φ−1(φ(S))=S\varphi^{-1}(\varphi(S)) = S for any S⊆GS \subseteq G as sets (standard fact: a function's inverse image of its own image of an injective map recovers the original set — indeed if g′∈φ−1(φ(S))g' \in \varphi^{-1}(\varphi(S)) then φ(g′)=φ(s)\varphi(g') = \varphi(s) for some s∈Ss\in S, and injectivity forces g′=s∈Sg'=s \in S; the reverse inclusion is immediate). Hence Ψ(Φ(S))=S\Psi(\Phi(S)) = S. Since φ\varphi is surjective, φ(φ−1(T))=T\varphi(\varphi^{-1}(T)) = T for any T⊆HT \subseteq H (surjectivity guarantees every element of TT has a preimage, which by definition lies in φ−1(T)\varphi^{-1}(T)). Hence Φ(Ψ(T))=T\Phi(\Psi(T)) = T. So Φ\Phi is a bijection between the subgroup lattices, with inverse Ψ\Psi. Containment is preserved: if S1≤S2≤GS_1 \le S_2 \le G then clearly φ(S1)⊆φ(S2)\varphi(S_1) \subseteq \varphi(S_2), and applying Ψ\Psi (order-preserving in the same way) to a containment in HH pulls back to a containment in GG; combined with bijectivity of Φ\Phi, containments correspond exactly (i.e. Φ(S1)≤Φ(S2)  ⟺  S1≤S2\Phi(S_1)\le \Phi(S_2) \iff S_1 \le S_2). Finally, for indices: if [G:S][G:S] is finite, the left cosets gSgS partition GG into [G:S][G:S] classes, and φ\varphi maps the coset gSgS bijectively onto the coset φ(g)φ(S)\varphi(g)\varphi(S) (bijectively because φ\varphi restricted to any subset is injective, being injective on all of GG; and φ(gs)=φ(g)φ(s)\varphi(gs) = \varphi(g)\varphi(s) shows the image is exactly φ(g)φ(S)\varphi(g)\varphi(S)). Distinct cosets g1S≠g2Sg_1 S \ne g_2 S map to distinct cosets φ(g1)φ(S)≠φ(g2)φ(S)\varphi(g_1)\varphi(S) \ne \varphi(g_2)\varphi(S): if these images were equal then φ(g1)−1φ(g2)∈φ(S)\varphi(g_1)^{-1}\varphi(g_2) \in \varphi(S), i.e. φ(g1−1g2)∈φ(S)\varphi(g_1^{-1}g_2) \in \varphi(S), and since φ\varphi is injective onto its image with φ−1(φ(S))=S\varphi^{-1}(\varphi(S)) = S as shown above, this gives g1−1g2∈Sg_1^{-1}g_2 \in S, i.e. g1S=g2Sg_1S = g_2S, a contradiction. Since φ\varphi is surjective onto HH, every coset of φ(S)\varphi(S) in HH arises this way. Hence the map gS↦φ(g)φ(S)gS \mapsto \varphi(g)\varphi(S) is a bijection between the coset spaces, so [G:S]=[H:φ(S)][G:S] = [H:\varphi(S)]. ■\blacksquare

∎

Worked examples

Example 1: (R,+)≅(R>0,×)(\mathbb{R}, +) \cong (\mathbb{R}_{>0}, \times). Let G=(R,+)G = (\mathbb{R}, +) (all reals under addition) and H=(R>0,×)H = (\mathbb{R}_{>0}, \times) (positive reals under multiplication). Define φ:G→H\varphi: G \to H by φ(x)=ex\varphi(x) = e^x. Homomorphism. For x,y∈Rx, y \in \mathbb{R}, φ(x+y)=ex+y=exey=φ(x)φ(y)\varphi(x+y) = e^{x+y} = e^x e^y = \varphi(x)\varphi(y), using the exponential law ex+y=exeye^{x+y}=e^xe^y (a fact about real exponentiation, not group theory). Injective. If φ(x)=φ(y)\varphi(x) = \varphi(y) then ex=eye^x = e^y; since ete^t is strictly increasing on R\mathbb{R}, this forces x=yx = y. Surjective. Given any r∈R>0r \in \mathbb{R}_{>0}, the real number x=ln⁡rx = \ln r satisfies φ(x)=eln⁡r=r\varphi(x) = e^{\ln r} = r (existence of ln⁡r\ln r for r>0r>0 is the standard fact that ete^t is a bijection R→R>0\mathbb{R} \to \mathbb{R}_{>0}). So φ\varphi is a bijective homomorphism, hence an isomorphism, and (R,+)≅(R>0,×)(\mathbb{R},+) \cong (\mathbb{R}_{>0}, \times). Note the asymmetry of this example: additive and multiplicative structures on entirely different underlying sets (all reals versus only positive reals) turn out to encode the same group. As a sanity check against part (b) of the theorem: both groups are abelian (matches (4)); both are torsion-free — every non-identity element has infinite order in (R,+)(\mathbb{R},+) (since nx=0⇒x=0nx=0 \Rightarrow x=0 for n≠0n\ne0), and correspondingly ex≠1e^x \ne 1 for x≠0x \ne 0 has (ex)n=enx≠1(e^x)^n = e^{nx}\ne 1 for n≠0n \ne 0 (matches (3)); neither is cyclic, since R\mathbb{R} is uncountable while a cyclic group generated by one element {ng:n∈Z}\{ng : n \in \mathbb{Z}\} is countable (matches (5), consistently — cyclicity transfers, and here both sides fail it).

Example 2: U(8)≅Z/2Z×Z/2ZU(8) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}, exhibited via an explicit non-obvious bijection. Let U(8)={1,3,5,7}U(8) = \{1,3,5,7\}, the group of units modulo 88 under multiplication mod 88, and let K=Z/2Z×Z/2Z={(0,0),(1,0),(0,1),(1,1)}K = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} = \{(0,0),(1,0),(0,1),(1,1)\} under componentwise addition mod 2. First compute the structure of U(8)U(8): 32=9≡13^2 = 9 \equiv 1, 52=25≡15^2=25\equiv 1, 72=49≡1(mod8)7^2 = 49 \equiv 1 \pmod 8. So every non-identity element of U(8)U(8) has order 22 — this is the Klein four-group pattern, matching KK where every non-identity element (1,0),(0,1),(1,1)(1,0),(0,1),(1,1) also has order 22 (e.g. (1,0)+(1,0)=(0,0)(1,0)+(1,0)=(0,0)). Define φ:U(8)→K\varphi: U(8) \to K by

φ(1)=(0,0),φ(3)=(1,0),φ(5)=(0,1),φ(7)=(1,1).\varphi(1) = (0,0), \quad \varphi(3) = (1,0), \quad \varphi(5) = (0,1), \quad \varphi(7) = (1,1).

This is manifestly a bijection between two 4-element sets. To check it is a homomorphism we must verify φ(ab  8)=φ(a)+φ(b)\varphi(ab \bmod 8) = \varphi(a)+\varphi(b) for all 99 non-identity ordered pairs beyond trivial identity cases (using that φ(1)\varphi(1) is the identity handles all pairs involving 11 automatically, by part (b)(1)'s converse check — actually we verify directly, not assuming the theorem):

  • 3⋅3=9≡13 \cdot 3 = 9 \equiv 1: need φ(3)+φ(3)=(1,0)+(1,0)=(0,0)=φ(1)\varphi(3)+\varphi(3) = (1,0)+(1,0) = (0,0) = \varphi(1). ✓.
  • 3⋅5=15≡73 \cdot 5 = 15 \equiv 7: need φ(3)+φ(5)=(1,0)+(0,1)=(1,1)=φ(7)\varphi(3)+\varphi(5) = (1,0)+(0,1) = (1,1) = \varphi(7). ✓.
  • 3⋅7=21≡53 \cdot 7 = 21 \equiv 5: need φ(3)+φ(7)=(1,0)+(1,1)=(0,1)=φ(5)\varphi(3)+\varphi(7) = (1,0)+(1,1)=(0,1) = \varphi(5). ✓.
  • 5⋅5=25≡15 \cdot 5 = 25 \equiv 1: need φ(5)+φ(5)=(0,1)+(0,1)=(0,0)=φ(1)\varphi(5)+\varphi(5) = (0,1)+(0,1) = (0,0) = \varphi(1). ✓.
  • 5⋅7=35≡35\cdot 7 = 35 \equiv 3: need φ(5)+φ(7)=(0,1)+(1,1)=(1,0)=φ(3)\varphi(5)+\varphi(7) = (0,1)+(1,1) = (1,0) = \varphi(3). ✓.
  • 7⋅7=49≡17 \cdot 7 = 49 \equiv 1: need φ(7)+φ(7)=(1,1)+(1,1)=(0,0)=φ(1)\varphi(7)+\varphi(7) = (1,1)+(1,1)=(0,0)=\varphi(1). ✓.

All cases check (multiplication in U(8)U(8) is commutative, so we needn't separately check reversed-order products). Hence φ\varphi is an isomorphism, and U(8)≅Z/2Z×Z/2ZU(8) \cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}. This example is instructive precisely because U(8)U(8) is not isomorphic to Z/4Z\mathbb{Z}/4\mathbb{Z}, even though both have order 44: Z/4Z\mathbb{Z}/4\mathbb{Z} has an element of order 44 (namely 11), but by our computation every non-identity element of U(8)U(8) has order 22. By part (b)(3) of the theorem, an isomorphism would have to match element orders exactly, so no isomorphism U(8)→Z/4ZU(8) \to \mathbb{Z}/4\mathbb{Z} can exist — the two groups of order 44 genuinely differ.

CounterexampleWhy bijectivity cannot be dropped from the definition

Drop the requirement that φ\varphi be a bijection, keeping only "homomorphism," and ask whether such a map alone should be called an isomorphism. It should not — here is why the theorem's conclusions collapse without bijectivity. Let G=(Z,+)G = (\mathbb{Z}, +) and H=(Z/4Z,+)H = (\mathbb{Z}/4\mathbb{Z}, +), and define φ:G→H\varphi: G \to H by φ(n)=n  4\varphi(n) = n \bmod 4. This is a homomorphism: φ(m+n)=(m+n)  4=(m  4)+(n  4)=φ(m)+φ(n)\varphi(m+n) = (m+n)\bmod 4 = (m \bmod 4) + (n \bmod 4) = \varphi(m)+\varphi(n) in Z/4Z\mathbb{Z}/4\mathbb{Z}, using that reduction mod 44 respects addition. It is surjective (every residue class is hit) but not injective: φ(0)=φ(4)=0\varphi(0) = \varphi(4) = 0, yet 0≠40 \ne 4 in Z\mathbb{Z}. Now every conclusion of part (b) that depended on bijectivity fails to be forced:

  • Order is not preserved. The element 1∈Z1 \in \mathbb{Z} has infinite order, but φ(1)=1∈Z/4Z\varphi(1) = 1 \in \mathbb{Z}/4\mathbb{Z} has order 44 (finite). A genuine isomorphism could never do this, by part (b)(3) — but a mere non-injective homomorphism routinely does.
  • Cyclicity claims break asymmetrically but not vacuously — a more serious failure appears with a different example. Take G=Z×ZG = \mathbb{Z}\times\mathbb{Z} (not cyclic — no single element generates it, since Z×Z\mathbb{Z}\times\mathbb{Z} requires two independent generators) and H=Z/5ZH = \mathbb{Z}/5\mathbb{Z} (cyclic, generated by 11). The map φ(m,n)=m+4n  5\varphi(m,n) = m + 4n \bmod 5 is a surjective homomorphism (check: φ((m1,n1)+(m2,n2))=φ(m1+m2,n1+n2)=(m1+m2)+4(n1+n2)  5=φ(m1,n1)+φ(m2,n2)\varphi((m_1,n_1)+(m_2,n_2)) = \varphi(m_1+m_2,n_1+n_2) = (m_1+m_2)+4(n_1+n_2) \bmod 5 = \varphi(m_1,n_1)+\varphi(m_2,n_2)), yet GG is not cyclic while HH is. If we mistakenly treated surjective homomorphisms as sufficient to conclude "GG is cyclic because HH is" we would derive a false statement.
  • Injectivity is what fails, since ker⁡φ={(m,n):m≡−4n≡n(mod5)}\ker\varphi = \{(m,n): m \equiv -4n \equiv n \pmod 5\} is infinite, so φ\varphi is very far from a bijection.

The moral: a homomorphism alone transports some information (e.g. surjective images of cyclic groups are cyclic — a fact provable directly and used elsewhere), but only a bijective homomorphism guarantees the two-sided transport of every structural property proven in part (b). Without an inverse homomorphism to run the argument backward — exactly what symmetry of ≅\cong required in the proof — order, cyclicity, and abelian-ness can each fail to transfer faithfully from domain to codomain or vice versa.

Scope, edge cases & common traps

SituationWhat to checkCommon pitfall
Finite groups, same orderCompare element-order spectra, abelian-nessAssuming same order ⇒\Rightarrow isomorphic (false: Z/4Z\mathbb{Z}/4\mathbb{Z} vs. U(8)U(8), both order 4)
Infinite groupsTorsion (element orders), cardinality, cyclicityAssuming a bijection of underlying sets is enough; must also respect the operation
Proving G≅HG \cong HExhibit explicit φ\varphi; verify homomorphism AND bijection, both directionsChecking homomorphism property on only a generating set without justifying it extends correctly
Proving G≇HG \not\cong HFind one invariant from part (b) that differsChecking only a few cases of a candidate map fails, then wrongly concluding no isomorphism exists at all
Automorphisms (G≅GG\cong G)Identity always works; look for othersAssuming the identity map is the only isomorphism G→GG\to G

The mathematician behind it

Origins · 1 minÉvariste Galois and how this was discovered
ÉG
Évariste Galois
1811–1832 · Early 19th-century France, theory of equations

Galois did not write down the modern definition of isomorphism, but his work on when two polynomial equations have "the same" solvability structure — via what we now call Galois groups and their relationships — planted the conceptual seed. The formal, set-theoretic language of isomorphism as a bijective structure-preserving map crystallized later, through the abstract group theory of Cayley, Jordan, and especially the abstract axiomatic tradition of Dedekind, Weber, and the early 20th-century algebraists (Noether, van der Waerden) who insisted on studying groups purely by their operation-table structure, independent of any particular realization.

History
- 1832: Galois's final letters, written the night before his fatal duel, describe correspondences between solvability of equations and the structure of associated permutation groups.
- 1854: Cayley gives an early abstract definition of a finite group via a multiplication table, implicitly treating groups up to relabeling.
- 1870: Jordan's Traité des substitutions systematizes permutation-group techniques, comparing groups by structural properties rather than by their concrete realization.
- 1893–1897: Hölder and others formalize the modern axiomatic notion of abstract group, making "isomorphism" a precise, checkable relation rather than an informal similarity.
- 1930: van der Waerden's Moderne Algebra, building on Noether's structural viewpoint, presents isomorphism as foundational to the entire subject, the form used ever since.
Is an isomorphism the same thing as an equality of groups?

No. Two isomorphic groups can have completely different underlying sets and different-looking operations; isomorphism says there is a structure-preserving bijection between them, not that they are literally the same set with the same operation symbol. For example (R,+)≅(R>0,×)(\mathbb{R},+) \cong (\mathbb{R}_{>0},\times), proved above via x↦exx \mapsto e^x, yet these are visibly different sets with different operations. What isomorphism guarantees is that every purely group-theoretic statement — about orders, subgroups, generating sets, commutativity — transfers perfectly between them. In practice, algebraists often abuse language and say "the group is Z/nZ\mathbb{Z}/n\mathbb{Z}" meaning "isomorphic to Z/nZ\mathbb{Z}/n\mathbb{Z}," since isomorphic groups are indistinguishable for all group-theoretic purposes.

Can a group be isomorphic to a proper subgroup of itself?

Yes, but only for infinite groups. For a finite group GG, any subgroup H<GH < G has ∣H∣<∣G∣|H| < |G| (strict, since HH is proper), so no bijection G→HG \to H can exist at all, let alone an isomorphism — finite sets of different sizes are never in bijection. For infinite groups this restriction vanishes: (Z,+)(\mathbb{Z},+) is isomorphic to its proper subgroup 2Z2\mathbb{Z} via φ(n)=2n\varphi(n) = 2n, which is a bijective homomorphism (φ(m+n)=2(m+n)=2m+2n=φ(m)+φ(n)\varphi(m+n)=2(m+n)=2m+2n=\varphi(m)+\varphi(n), and φ\varphi is clearly a bijection onto 2Z2\mathbb{Z}). This is a classic instance of Hilbert's Hotel-style phenomena unique to infinite structures.

Does isomorphism preserve normality of subgroups?

Yes. If φ:G→H\varphi: G \to H is an isomorphism and N⊴GN \trianglelefteq G is a normal subgroup, then φ(N)⊴H\varphi(N) \trianglelefteq H. This follows because normality means gNg−1=NgNg^{-1} = N for all g∈Gg \in G, and applying φ\varphi (using the homomorphism property and that φ\varphi is a bijection matching inverses, as proved in part (b)(2) above) gives φ(g)φ(N)φ(g)−1=φ(N)\varphi(g)\varphi(N)\varphi(g)^{-1} = \varphi(N) for all φ(g)∈H\varphi(g) \in H, which covers all of HH by surjectivity. This is one more instance of the general principle that isomorphisms preserve every property expressible purely via the group operation, extending the list proved in the central theorem.

How do I actually prove two specific groups are isomorphic, in practice?

The standard method is to (1) guess a natural correspondence, often suggested by comparing generators or structural data such as order and cyclicity, (2) define an explicit map φ\varphi, (3) verify the homomorphism identity φ(ab)=φ(a)φ(b)\varphi(ab)=\varphi(a)\varphi(b) carefully — often it suffices to check it on generators if you also justify the extension, and (4) verify φ\varphi is a bijection, typically by exhibiting an explicit inverse or via a counting argument on finite sets (injective + same finite cardinality implies bijective). Example 2 above illustrates this fully worked out for U(8)≅Z/2Z×Z/2ZU(8) \cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}.

How do I prove two groups are NOT isomorphic without checking every map?

Use invariants proved in part (b) of the theorem: compare orders, abelian-ness, cyclicity, or — most powerfully for finite groups — the multiset of element orders. If any of these differ, no isomorphism can exist, since an isomorphism would force them to agree exactly. This is far more efficient than attempting to rule out every bijection individually, which is combinatorially infeasible even for moderately sized finite groups (there are n!n! bijections between two sets of size nn).

Is every homomorphism between groups of the same finite order automatically an isomorphism?

No — this is a subtle trap. A homomorphism between finite groups of the same order is an isomorphism if and only if it is injective (equivalently, surjective), because for maps between finite sets of equal size, injectivity and surjectivity are equivalent, and either one plus homomorphism gives bijectivity. But a homomorphism need not be injective at all: the trivial map sending every element of GG to eHe_H is always a homomorphism regardless of ∣G∣=∣H∣|G|=|H|, yet it is almost never a bijection (unless both groups are trivial). So same order alone guarantees nothing; you must separately verify injectivity or surjectivity of the specific map in hand.

What is the relationship between isomorphisms and automorphisms?

An automorphism is precisely an isomorphism from a group GG to itself. By the theorem's part (a), the identity map is always an automorphism, and the set of all automorphisms of GG, denoted Aut⁡(G)\operatorname{Aut}(G), forms a group under composition (associativity from function composition, identity from id⁡G\operatorname{id}_G, inverses from the symmetric part of the proof above showing inverses of isomorphisms are isomorphisms). Automorphism groups measure the internal symmetry of GG and play a central role in constructions such as semidirect products, extending the ideas here toward Direct Products and beyond.

Before this

Where this leads