Groups often arrive wearing costumes. The integers under addition, rotations of a polygon, invertible matrices, and permutations all look different because their elements have different names and live in different ambient worlds. Group isomorphism asks a sharper question: after forgetting the labels and keeping only the operation, are two groups the same structure?

A plain bijection cannot answer this question. A bijection can rename the elements of a group in a way that destroys multiplication. A plain homomorphism cannot answer it either, because it may collapse distinct elements or miss part of the target. The right notion of sameness must preserve the operation and lose no information.

[example: Additive Residues and Fourth Roots of Unity] Here $\mathbb{Z}/4\mathbb{Z}$ denotes the four residue classes of integers modulo $4$, written $\bar{0},\bar{1},\bar{2},\bar{3}$, where $\bar{k}=\bar{\ell}$ means that $k-\ell$ is divisible by $4$. Also, $\mathbb{C}^{\times}$ denotes the group of nonzero complex numbers under multiplication, and $\subsetneq$ means "proper subset." Let $G=(\mathbb{Z}/4\mathbb{Z},+)$ and let $H=\{1,i,-1,-i\}\subsetneq \mathbb{C}^{\times}$ under multiplication. Define $\varphi:G\to H$ by $\varphi(\bar{k})=i^k$. This is well-defined: if $\bar{k}=\bar{\ell}$ in $\mathbb{Z}/4\mathbb{Z}$, then $k-\ell=4m$ for some $m\in\mathbb{Z}$, so \begin{align*} i^k=i^{\ell+4m}=i^\ell(i^4)^m=i^\ell\cdot 1^m=i^\ell. \end{align*} We show that $\varphi$ preserves the group operation. For $\bar{a},\bar{b}\in \mathbb{Z}/4\mathbb{Z}$, addition of residue classes gives $\bar{a}+\bar{b}=\overline{a+b}$, and therefore \begin{align*} \varphi(\bar{a}+\bar{b})=\varphi(\overline{a+b})=i^{a+b}. \end{align*} By the exponent law in $\mathbb{C}^{\times}$, \begin{align*} i^{a+b}=i^a i^b=\varphi(\bar{a})\varphi(\bar{b}). \end{align*} Thus $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism). The values on the four residue classes are \begin{align*} \varphi(\bar{0})=i^0=1,\quad \varphi(\bar{1})=i,\quad \varphi(\bar{2})=-1,\quad \varphi(\bar{3})=-i. \end{align*} These are exactly the four elements of $H$, and they are pairwise distinct, so $\varphi$ is both surjective and injective. Hence $\varphi$ is a bijective homomorphism from $G$ to $H$, so the additive cycle $\bar{0},\bar{1},\bar{2},\bar{3}$ and the multiplicative cycle $1,i,-1,-i$ are the same group after this consistent relabelling. [/example]

This example is the whole story in miniature. The notation changed from addition to multiplication, and the elements changed from residue classes to complex numbers, yet the multiplication table survived exactly. Group isomorphism formalizes precisely this kind of structural identity.

## Homomorphisms as Structure-Preserving Maps

The parent notion is a [group](/page/Group): a set with an associative binary operation, an identity element, and inverses. To compare two groups, we first need maps that respect the operations. Without this condition, a function between the underlying sets has no reason to see the group structure.

[definition: Group Homomorphism] Let $(G, \cdot)$ and $(H, *)$ be groups. A group homomorphism from $G$ to $H$ is a function $\varphi: G \to H$ such that \begin{align*} \varphi(g_1 \cdot g_2) = \varphi(g_1) * \varphi(g_2) \end{align*} for all $g_1, g_2 \in G$. [/definition]

A homomorphism respects products, but it may still forget information. The quotient map $\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ collapses many integers to one residue class, while the inclusion $\mathbb{Z} \to \mathbb{Q}$ misses most rational numbers. This is why sameness of groups cannot mean merely "there is a homomorphism between them"; the comparison must also be reversible at the level of elements.

## Definition

The central definition combines these two demands. The map must preserve the operation, and it must be a complete relabelling of the underlying set. That is exactly the point at which two groups have the same multiplication table, possibly written with different symbols.

[definition: Group Isomorphism] Let $(G, \cdot)$ and $(H, *)$ be groups. A group isomorphism from $G$ to $H$ is a bijective group homomorphism $\varphi: G \to H$. [/definition]

The word "bijective" is doing structural work. Injectivity says different elements of $G$ remain different after relabelling; surjectivity says every element of $H$ receives a label from $G$. The homomorphism condition then says the operation is transported faithfully.

## Structure Preserved by Isomorphism

The point of an isomorphism is that every statement written only in terms of the group operation remains true after translation. Before using this as a test, we need the corresponding relation between groups themselves, rather than only the maps that witness the relation.

[definition: Isomorphic Groups] Let $G$ and $H$ be groups. The groups $G$ and $H$ are isomorphic, written $G \cong H$, if there exists a group isomorphism $\varphi: G \to H$. [/definition]

The notation $G \cong H$ records existence, not a particular choice. There may be many isomorphisms between the same two groups, and choosing one can matter when transporting specific elements or subgroups. The question remaining is whether this relation can always be reversed by another group isomorphism.

[quotetheorem:770]

This makes isomorphism symmetric: if $G$ can be translated into $H$, then $H$ can be translated back into $G$. Identities, inverses, powers, generated subgroups, commutation, and element orders all survive this reversible translation. This section turns that guiding principle into usable tests.

### Identity, Inverses, and Powers

The homomorphism law only mentions products, but the identity element and inverses are determined by products. Without a separate check, it is not obvious that preserving multiplication also forces a homomorphism to preserve the distinguished element $1_G$ and the inverse operation. This is the first obstruction to treating an isomorphism as a complete relabelling of a group: the relabelling must respect the pieces of the group law that are only implicit in the product rule.