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. [quotetheorem:4981] Repeated multiplication is how a single element reveals its internal cycle. Even after identities and inverses are controlled, there is still a compatibility question for longer products: applying an isomorphism before or after taking repeated powers should give the same result. This matters because powers are the language used to describe generated subgroups and element orders. [quotetheorem:9502] The most compact way to record the cycle generated by an element is to ask when its powers return to the identity. Some elements return after finitely many steps, while others never do, and an invariant has to record both possibilities. To compare elements across groups without referring to their names, we need a precise term for this return time. [definition: Order of an Element] Let $G$ be a group. The order function of $G$ is the function $\operatorname{ord}: G \to \mathbb{N} \cup \{\infty\}$ such that, for each $g \in G$, $\operatorname{ord}(g)$ is the smallest positive integer $n \ge 1$ satisfying $g^n = 1_G$, if such an $n$ exists, and $\operatorname{ord}(g)=\infty$ otherwise. [/definition] Element order is not a feature of the element's name. It is a feature of the way the element multiplies with itself. To use element order as a comparison tool, the obstruction is a possible change in cycle length under relabelling: an isomorphism might a priori send an element with one return time to an element with another. Since the equation $g^n=1_G$ is expressed only through powers and the identity, the next structural invariant to check is whether isomorphisms preserve the smallest such exponent. [quotetheorem:9503] A mismatch in element orders rules out an isomorphism immediately. This gives a fast way to see that two groups of the same size may still have different algebraic structure. [example: Distinguishing $\mathbb{Z}/4\mathbb{Z}$ from the Klein Four Group] Let $G=\mathbb{Z}/4\mathbb{Z}$ under addition, and let $V=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ under componentwise addition. In $G$, the identity is $\bar{0}$, and the successive sums of $\bar{1}$ are \begin{align*} 1\bar{1}=\bar{1},\quad 2\bar{1}=\bar{1}+\bar{1}=\bar{2},\quad 3\bar{1}=\bar{3},\quad 4\bar{1}=\bar{4}=\bar{0}. \end{align*} The first positive multiple of $\bar{1}$ equal to $\bar{0}$ is therefore $4$, so $\operatorname{ord}(\bar{1})=4$. In $V$, the identity is $(\bar{0},\bar{0})$. The elements are $(\bar{0},\bar{0})$, $(\bar{1},\bar{0})$, $(\bar{0},\bar{1})$, and $(\bar{1},\bar{1})$. For any $(a,b)\in V$, componentwise addition gives \begin{align*} (a,b)+(a,b)=(a+a,b+b)=(\bar{0},\bar{0}), \end{align*} because $a+a=\bar{0}$ and $b+b=\bar{0}$ in $\mathbb{Z}/2\mathbb{Z}$. Thus each non-identity element of $V$ has order $2$: it is not $(\bar{0},\bar{0})$, but adding it to itself gives $(\bar{0},\bar{0})$. By *[Isomorphisms Preserve Element Orders](/theorems/9503)*, an isomorphism would send an element of order $4$ in $G$ to an element of order $4$ in $V$. Since $G$ has such an element and $V$ has none, $G\not\cong V$. [/example] The example shows that cardinality is too crude an invariant. Both groups have four elements, and both are abelian, but their cyclic behaviour differs. ### Subgroups and Generated Structure A group is not only a collection of elements; it contains smaller groups inside it. If a map respects the group operation, then it should relate the pieces generated by chosen elements in the domain to the pieces generated by their images in the target. To state this comparison, we first need to name the subgroup generated by a set. [definition: Generated Subgroup] Let $G$ be a group and let $S \subset G$. The subgroup generated by $S$, denoted $\langle S \rangle$, is the smallest subgroup of $G$ containing $S$. [/definition] Generation asks what can be built from a chosen set using the group operation and inverses. For an arbitrary homomorphism, some elements may collapse and the target may not be fully reached, but every product and inverse built from $S$ is still sent to the corresponding product and inverse built from the image of $S$. Thus the next invariant begins at the level of homomorphisms: it compares the image of the subgroup generated by $S$ with the subgroup generated by the image of $S$. [quotetheorem:9504] The special case where one element generates the whole group is central enough to name. In that situation, every element of the group is forced by the repeated powers of a single element, so many questions about the whole group reduce to questions about one generator. We need a term for groups with this one-generator description before using it as a classification test. [definition: Cyclic Group] A group $G$ is cyclic if there exists $g \in G$ such that $G = \langle g \rangle$. [/definition] For cyclic groups, once the image of a generator is chosen, the entire homomorphism is forced. This makes cyclic groups the first place where the classification question below can be answered completely. [quotetheorem:9505] The theorem says that a [cyclic group](/page/Cyclic%20Group) is determined up to isomorphism by its size. It does not say the isomorphism is unique, because different choices of generator give different relabellings. [example: Choosing a Generator Determines the Isomorphism] Let $G=\langle g\rangle$ be cyclic of order $6$, so $g^6=1_G$ and no positive power $g^m$ with $1\le m<6$ equals $1_G$. Define $\varphi:\mathbb{Z}/6\mathbb{Z}\to G$ by $\varphi(\bar{k})=g^k$. This is well-defined: if $\bar{k}=\bar{\ell}$, then $k-\ell=6q$ for some $q\in\mathbb{Z}$, and hence \begin{align*} g^k=g^{\ell+6q}=g^\ell(g^6)^q=g^\ell 1_G^q=g^\ell. \end{align*} For $\bar{a},\bar{b}\in\mathbb{Z}/6\mathbb{Z}$, addition of residue classes gives $\bar{a}+\bar{b}=\overline{a+b}$, so \begin{align*} \varphi(\bar{a}+\bar{b})=\varphi(\overline{a+b})=g^{a+b}=g^ag^b=\varphi(\bar{a})\varphi(\bar{b}). \end{align*} Thus $\varphi$ is a homomorphism. Its values on the six residue classes are \begin{align*} \varphi(\bar{0})=1_G,\quad \varphi(\bar{1})=g,\quad \varphi(\bar{2})=g^2,\quad \varphi(\bar{3})=g^3,\quad \varphi(\bar{4})=g^4,\quad \varphi(\bar{5})=g^5. \end{align*} These six elements are all distinct, because $g^r=g^s$ with $0\le r<s\le 5$ would imply $g^{s-r}=1_G$ with $1\le s-r\le 5$, contradicting that $g$ has order $6$. Since $G=\langle g\rangle=\{1_G,g,g^2,g^3,g^4,g^5\}$, the map is also surjective. Therefore $\varphi$ is an isomorphism. Now set $h=g^5$. Its powers are \begin{align*} h^0=1_G,\quad h^1=g^5,\quad h^2=g^{10}=g^4,\quad h^3=g^{15}=g^3,\quad h^4=g^{20}=g^2,\quad h^5=g^{25}=g. \end{align*} These are again exactly the six elements of $G$, so $h$ is a generator, and the same argument shows that $\bar{k}\mapsto h^k$ is another isomorphism $\mathbb{Z}/6\mathbb{Z}\to G$. By contrast, if $a=g^2$, then \begin{align*} a^1=g^2,\quad a^2=g^4,\quad a^3=g^6=1_G. \end{align*} Also $a\ne 1_G$ and $a^2\ne 1_G$, since otherwise $g^2=1_G$ or $g^4=1_G$ would contradict the order of $g$. Hence $a$ has order $3$, and the image of $\bar{k}\mapsto a^k$ is only $\{1_G,g^2,g^4\}$. It misses $g$, so it is not an isomorphism onto $G$. Thus a generator may be chosen in more than one way, but choosing a non-generator cannot produce the full cyclic group. [/example] The freedom is real but limited. A generator may be sent to any generator, and after that choice every element has a forced image. ## Constructing and Recognizing Isomorphisms To prove that two groups are isomorphic, one usually constructs a map and checks two things. First, it respects the operation. Second, it is bijective. The best way to check the second condition often depends on the group: direct counting, generators, kernels, images, or explicit inverse maps. ### Presentations and Relations Some groups are easier to recognize from a recipe than from a multiplication table. A small list of generators and relations can encode a large group: the generators are the basic moves, and the relations are the equations those moves must satisfy. For isomorphism arguments, the point is not the full construction of presentations, but the practical rule they give for building maps. [definition: Presentation of a Group] For this page, a presentation of a group means a description by generators and relations, written $\langle S \mid R \rangle$. The symbols in $S$ are named generators, and the relations in $R$ are equations involving those generators that are required to hold in the group. A group with such a presentation is generated by the listed generators, subject to the listed equations. [/definition] Presentations are useful for constructing homomorphisms because a map out of a group described this way is determined once the images of the generators are chosen. The necessary check is that the chosen images satisfy the same defining relations in the target group; then the recipe respects all products forced by those relations. If the chosen images also generate the target, the resulting homomorphism is a strong candidate for an isomorphism. [example: Recognizing the Dihedral Group of Order $6$] Let $D_6$ be the group of symmetries of an equilateral triangle. Label the vertices $1,2,3$ in counterclockwise order. Let $r$ be rotation by $2\pi/3$, so $r$ sends $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$, and let $s$ be the reflection fixing vertex $3$ and swapping vertices $1$ and $2$. Thus the action on vertices gives \begin{align*} r=(123) \quad \text{and} \quad s=(12). \end{align*} The rotation relation is visible from following vertices through three rotations: \begin{align*} r^3(1)=1,\quad r^3(2)=2,\quad r^3(3)=3. \end{align*} Hence $r^3=1$. Similarly, $s$ swaps $1$ and $2$ twice and fixes $3$ both times, so \begin{align*} s^2(1)=1,\quad s^2(2)=2,\quad s^2(3)=3. \end{align*} Hence $s^2=1$. Now compute the conjugation relation on each vertex, applying the rightmost permutation first. Since $r^{-1}=(132)$, we compare $srs$ with $r^{-1}$. For vertex $1$, \begin{align*} srs(1)=sr(2)=s(3)=3=r^{-1}(1). \end{align*} For vertex $2$, \begin{align*} srs(2)=sr(1)=s(2)=1=r^{-1}(2). \end{align*} For vertex $3$, \begin{align*} srs(3)=sr(3)=s(1)=2=r^{-1}(3). \end{align*} Thus $srs=r^{-1}$. The [symmetric group](/page/Symmetric%20Group) $S_3$ is generated by $\rho=(123)$ and $\sigma=(12)$. These satisfy the same relations: \begin{align*} \rho^3=1,\quad \sigma^2=1,\quad \sigma\rho\sigma=\rho^{-1}. \end{align*} The assignment $r\mapsto \rho$ and $s\mapsto \sigma$ therefore sends the defining products $r^3$, $s^2$, and $srsr$ to the identity in $S_3$. By the generator-and-relation principle just described, this assignment respects the defining rules for multiplying triangle symmetries, so it defines a homomorphism $\varphi:D_6\to S_3$ with $\varphi(r)=\rho$ and $\varphi(s)=\sigma$. The six symmetries of the triangle are \begin{align*} 1,\quad r,\quad r^2,\quad s,\quad rs,\quad r^2s. \end{align*} Their images are \begin{align*} 1,\quad (123),\quad (132),\quad (12),\quad (123)(12),\quad (132)(12). \end{align*} The last two products are \begin{align*} (123)(12)=(13). \end{align*} and \begin{align*} (132)(12)=(23). \end{align*} So the image contains \begin{align*} 1,\quad (123),\quad (132),\quad (12),\quad (13),\quad (23), \end{align*} which are all six elements of $S_3$. Hence $\varphi$ is surjective. Since $D_6$ also has exactly six elements, a surjective function $D_6\to S_3$ must be injective as well: six distinct elements in the domain map onto six elements in the codomain, so no two domain elements can share an image. Therefore $\varphi$ is a bijective homomorphism, so $D_6\cong S_3$. The triangle symmetries and the permutations of its three vertices are the same group structure written in two different languages. [/example] This is a recurring strategy: recognize a group not by the names of its elements, but by the relations among a small set of generators. ### Kernel and Image Tests When a homomorphism is already available, the remaining question is whether it loses information or misses elements. The kernel measures the first failure, and the image measures the second. [definition: Kernel of a Group Homomorphism] Let $G$ and $H$ be groups, and let $\varphi: G \to H$ be a group homomorphism. The kernel of $\varphi$ is \begin{align*} \ker \varphi = \{g \in G : \varphi(g) = 1_H\}. \end{align*} [/definition] A homomorphism with a non-identity element in its kernel has collapsed that element to the same value as $1_G$. That detects loss of information in the domain, but it says nothing about whether every element of the target is actually hit. To measure this second possible failure of being an isomorphism, we need to record the subset of the target that the homomorphism reaches. [definition: Image of a Group Homomorphism] Let $G$ and $H$ be groups, and let $\varphi: G \to H$ be a group homomorphism. The image of $\varphi$ is \begin{align*} \operatorname{im} \varphi = \{\varphi(g) : g \in G\}. \end{align*} [/definition] Together, kernel and image turn bijectivity into algebraic data. The practical problem is that checking a homomorphism directly for one-to-one and onto behavior can be awkward, while kernels and images are often computable from formulas. The isomorphism test below translates the two set-theoretic conditions into these algebraic measurements. [quotetheorem:9506] For finite groups of equal size, injectivity and surjectivity are not independent. The obstruction is simpler than in the general case: a one-to-one map between two finite sets of the same cardinality cannot miss a target element, and an onto map cannot identify two domain elements. This lets the isomorphism problem be reduced to checking only one half of bijectivity. [quotetheorem:9507] The two possible failures of bijectivity are different in practice. Seeing both side by side helps prevent the common mistake of treating every useful homomorphism as an isomorphism. [example: Homomorphisms That Are Not Isomorphisms] Consider the reduction map $\pi:\mathbb{Z}\to \mathbb{Z}/6\mathbb{Z}$ defined by $\pi(k)=\bar{k}$, with both groups written additively. For $a,b\in \mathbb{Z}$, addition of residue classes gives \begin{align*} \pi(a+b)=\overline{a+b}=\bar{a}+\bar{b}=\pi(a)+\pi(b). \end{align*} Thus $\pi$ is a group homomorphism. It is surjective because every element of $\mathbb{Z}/6\mathbb{Z}$ has the form $\bar{r}$ for some $r\in\mathbb{Z}$, and then $\pi(r)=\bar{r}$. Its kernel is \begin{align*} \ker \pi=\{k\in\mathbb{Z}:\pi(k)=\bar{0}\}=\{k\in\mathbb{Z}:\bar{k}=\bar{0}\}. \end{align*} The equality $\bar{k}=\bar{0}$ in $\mathbb{Z}/6\mathbb{Z}$ means exactly that $6$ divides $k$, so \begin{align*} \ker \pi=\{6m:m\in\mathbb{Z}\}=6\mathbb{Z}. \end{align*} Since $6\in \ker\pi$ and $6\ne 0$, the map is not injective; equivalently, $\pi(0)=\bar{0}=\pi(6)$ but $0\ne 6$. Now consider the inclusion map $\iota:\mathbb{Z}\to\mathbb{Q}$ defined by $\iota(k)=k$, again under addition. For $a,b\in\mathbb{Z}$, \begin{align*} \iota(a+b)=a+b=\iota(a)+\iota(b), \end{align*} so $\iota$ is a group homomorphism. If $\iota(a)=\iota(b)$, then $a=b$ as rational numbers, hence $a=b$ as integers; therefore $\iota$ is injective. It is not surjective, because every value of $\iota$ is an integer, while $\frac{1}{2}\in\mathbb{Q}$ is not an integer. Thus $\frac{1}{2}\notin \operatorname{im}\iota$. The first map reaches the whole target but collapses distinct integers modulo $6$, while the second map loses no information but misses most rational numbers. Both preserve addition, yet neither is an isomorphism. [/example] These examples show why bijectivity is part of the definition rather than a cosmetic add-on. A homomorphism compares groups; an isomorphism identifies them. ## Automorphisms and Internal Symmetry An isomorphism from $G$ to $H$ compares two groups. An isomorphism from $G$ to itself records a symmetry of the group structure. These self-relabellings are often more interesting than the identity map. [definition: Group Automorphism] Let $G$ be a group. A group automorphism of $G$ is a group isomorphism $\varphi: G \to G$. [/definition] Automorphisms can be composed, and the inverse of an automorphism is again an automorphism. This means all self-isomorphisms of $G$ carry their own group structure, motivating the definition below. [definition: Automorphism Group] Let $G$ be a group. The automorphism group of $G$, denoted $\operatorname{Aut}(G)$, is the group of all automorphisms of $G$ under composition. [/definition] For a cyclic group, every automorphism is determined by where a generator goes. The constraint is that the chosen image must still generate the whole group; otherwise the map forced by that choice cannot be onto. In a finite cyclic group, this turns the automorphism question into the arithmetic question of which powers of a generator are again generators. [quotetheorem:9508] The automorphism group still leaves open a source question: which self-symmetries arise from the multiplication already present inside $G$? In a non-abelian group, multiplying by a fixed element on the left and its inverse on the right can move elements while respecting their multiplication relationships. This internal conjugation construction is important enough to name because it produces automorphisms from elements of the group itself. [definition: Inner Automorphism] Let $G$ be a group and let $g \in G$. The inner automorphism associated to $g$ is the function $\iota_g: G \to G$ defined by $\iota_g(x)=gxg^{-1}$ for every $x \in G$. [/definition] The definition gives a formula, but the formula still has to earn its place in $\operatorname{Aut}(G)$. The obstruction is that $x\mapsto gxg^{-1}$ could merely be a rearrangement of symbols unless it is known to preserve products and be bijective. The next result supplies exactly that verification, showing that conjugation by a fixed group element is a genuine automorphism. [quotetheorem:4984] Inner automorphisms are invisible in abelian groups because $gxg^{-1}=x$ for every $x \in G$. In non-abelian groups, they encode the way a group acts on itself by conjugation. [example: Automorphisms of $S_3$ by Conjugation] Let $g\in S_3$, and write products as composition from right to left. We first compute the conjugate of the transposition $(12)$ by evaluating it on each possible input. Since $g$ is a bijection, $g(1)$ and $g(2)$ are distinct. For the first two points, \begin{align*} g(12)g^{-1}(g(1))=g(12)(1)=g(2). \end{align*} and \begin{align*} g(12)g^{-1}(g(2))=g(12)(2)=g(1). \end{align*} If $y\in\{1,2,3\}$ is different from both $g(1)$ and $g(2)$, then $g^{-1}(y)$ is different from both $1$ and $2$, so $(12)$ fixes $g^{-1}(y)$. Hence \begin{align*} g(12)g^{-1}(y)=g(g^{-1}(y))=y. \end{align*} Thus $g(12)g^{-1}$ swaps $g(1)$ and $g(2)$ and fixes the remaining point, so \begin{align*} g(12)g^{-1}=(g(1)\ g(2)). \end{align*} Therefore every conjugate of a transposition is again a transposition. The same calculation shows what happens to a $3$-cycle. For $c=(123)$, \begin{align*} gcg^{-1}(g(1))=gc(1)=g(2). \end{align*} and \begin{align*} gcg^{-1}(g(2))=gc(2)=g(3). \end{align*} and \begin{align*} gcg^{-1}(g(3))=gc(3)=g(1). \end{align*} Hence \begin{align*} g(123)g^{-1}=(g(1)\ g(2)\ g(3)). \end{align*} Since $g(1),g(2),g(3)$ are exactly the three symbols $1,2,3$ in some order, this conjugate is one of the two $3$-cycles in $S_3$. So conjugation in $S_3$ sends the set of three transpositions to itself and sends the set of two $3$-cycles to itself. This is symmetry of the multiplication structure of $S_3$, not an added geometric picture. [/example] Automorphisms make the distinction between equality and isomorphism vivid. The group has not changed, but its elements have been renamed in a way that the operation cannot detect. ## Invariants and Non-Isomorphism Many group theory arguments prove that two groups are not isomorphic. The strategy is to find a feature that any isomorphism would preserve and then show that the two groups disagree on that feature. [definition: Group Isomorphism Invariant] Let $\mathcal{C}$ be a collection of groups. A group isomorphism invariant on $\mathcal{C}$ is a function $I: \mathcal{C} \to X$, for some set $X$, such that $I(G)=I(H)$ whenever $G,H \in \mathcal{C}$ and $G \cong H$. [/definition] Invariants are useful because failed equality of invariants proves non-isomorphism. The most common elementary invariants are the size of the group, whether multiplication commutes, whether one element generates the whole group, and how many elements have each possible order; the next theorem gathers these tests. [quotetheorem:9509] The theorem is a toolbox for ruling out isomorphisms. A single mismatch is enough, although matching several invariants still may not prove that an isomorphism exists. [example: Separating Groups of Order $6$] The group $\mathbb{Z}/6\mathbb{Z}$ has the six residue classes $\bar{0},\bar{1},\bar{2},\bar{3},\bar{4},\bar{5}$, and $S_3$ has the six permutations of $\{1,2,3\}$. We show that they are not isomorphic by comparing commutativity. First, $\mathbb{Z}/6\mathbb{Z}$ is abelian. For any $\bar{a},\bar{b}\in \mathbb{Z}/6\mathbb{Z}$, \begin{align*} \bar{a}+\bar{b}=\overline{a+b}. \end{align*} Since addition in $\mathbb{Z}$ is commutative, $a+b=b+a$, so \begin{align*} \overline{a+b}=\overline{b+a}=\bar{b}+\bar{a}. \end{align*} Thus $\bar{a}+\bar{b}=\bar{b}+\bar{a}$ for all residue classes. Now compute two products in $S_3$, using right-to-left composition. For $(12)(23)$, the element $1$ is sent by $(23)$ to $1$ and then by $(12)$ to $2$; the element $2$ is sent by $(23)$ to $3$ and then fixed by $(12)$; the element $3$ is sent by $(23)$ to $2$ and then by $(12)$ to $1$. Hence \begin{align*} (12)(23)=(123). \end{align*} For $(23)(12)$, the element $1$ is sent by $(12)$ to $2$ and then by $(23)$ to $3$; the element $2$ is sent by $(12)$ to $1$ and then fixed by $(23)$; the element $3$ is fixed by $(12)$ and then sent by $(23)$ to $2$. Hence \begin{align*} (23)(12)=(132). \end{align*} Since $(123)\ne (132)$, the two elements $(12)$ and $(23)$ do not commute, so $S_3$ is not abelian. If an isomorphism $\varphi:\mathbb{Z}/6\mathbb{Z}\to S_3$ existed, then every $x,y\in S_3$ would have the form $x=\varphi(\bar{a})$ and $y=\varphi(\bar{b})$ by surjectivity. Using the homomorphism property and commutativity in $\mathbb{Z}/6\mathbb{Z}$, \begin{align*} xy=\varphi(\bar{a})\varphi(\bar{b})=\varphi(\bar{a}+\bar{b})=\varphi(\bar{b}+\bar{a})=\varphi(\bar{b})\varphi(\bar{a})=yx. \end{align*} This would make $S_3$ abelian, contradicting the calculation above. Therefore $\mathbb{Z}/6\mathbb{Z}\not\cong S_3$. [/example] Sometimes elementary invariants must be refined. Two non-abelian groups of the same order may still be separated by counting elements of each order. [example: Quaternion and Dihedral Groups] Let $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$, with multiplication determined by \begin{align*} i^2=j^2=k^2=-1. \end{align*} Then $1$ has order $1$, and \begin{align*} (-1)^2=1. \end{align*} Since $-1\ne 1$, the element $-1$ has order $2$. For $x\in\{i,j,k\}$, \begin{align*} x^2=-1. \end{align*} Thus $x^2\ne 1$, while \begin{align*} x^4=(x^2)^2=(-1)^2=1. \end{align*} So $i,j,k$ have order $4$. The negatives have the same square: \begin{align*} (-x)^2=(-1)^2x^2=x^2=-1. \end{align*} Hence $(-x)^2\ne 1$ and $(-x)^4=((-x)^2)^2=(-1)^2=1$, so $-i,-j,-k$ also have order $4$. Therefore $Q_8$ has exactly one element of order $2$, namely $-1$. Now write $D_8$ as the symmetry group of a square, with $r$ the rotation by $90^\circ$ and $s$ a reflection. Its elements are \begin{align*} 1,\ r,\ r^2,\ r^3,\ s,\ rs,\ r^2s,\ r^3s. \end{align*} The rotation satisfies $r^4=1$, so \begin{align*} (r^2)^2=r^4=1. \end{align*} Since $r^2$ is the rotation by $180^\circ$, it is not the identity, so $r^2$ has order $2$. Also $s^2=1$, and each $r^a s$ is a reflection of the square, so it is not the identity. Using the reflection relation $sr^a=r^{-a}s$, \begin{align*} (r^a s)^2=r^a s r^a s=r^a r^{-a}s^2=1. \end{align*} Thus $s,rs,r^2s,r^3s$ all have order $2$. Hence $D_8$ has at least five elements of order $2$. An isomorphism preserves the number of elements of each finite order by *Basic Group Isomorphism Invariants*. Since $Q_8$ has exactly one element of order $2$ and $D_8$ has more than one, the two groups cannot be isomorphic. Therefore $Q_8\not\cong D_8$. [/example] Commutation with every element is another structural feature that survives isomorphism. The issue is that some elements may behave as though the group were abelian even when the whole group is not: they commute with every possible partner. To use this collection as an invariant, we need a precise name for the subgroup of universally commuting elements. [definition: Centre of a Group] Let $G$ be a group. The centre of $G$ is \begin{align*} Z(G) = \{g \in G : gx = xg \text{ for all } x \in G\}. \end{align*} [/definition] The centre is defined using only the multiplication law, but invariance is not automatic from the notation alone. The obstruction is that centrality quantifies over every element of the group, so after applying an isomorphism one must compare commuting with all elements of the target group, not only with the images already named. This is the point at which centrality becomes a genuine isomorphism invariant and links naturally with conjugation. [quotetheorem:9510] Centres are especially useful in non-abelian group theory. Through the conjugation action $G \to \operatorname{Aut}(G)$, the centre measures the part of the group that is invisible to inner automorphisms; this viewpoint is developed further when inner automorphism groups are studied in their own right. ## Isomorphism as an Equivalence Relation Classification in algebra usually means classification up to isomorphism. For this to behave like a sensible notion of sameness, isomorphism must have the formal properties of equality: every group is isomorphic to itself, isomorphisms reverse, and isomorphisms compose. [definition: Equivalence Relation] Let $X$ be a set. An [equivalence relation](/page/Equivalence%20Relation) on $X$ is a relation $\sim$ such that for all $x,y,z \in X$, the relation is reflexive, symmetric, and transitive. [/definition] Equivalence relations are the language of classification. They divide a universe of objects into classes whose members are considered the same for the chosen purpose. For groups, the obstruction is that isomorphism is defined by the existence of a structure-preserving bijection, and it must still behave consistently under doing nothing, reversing a comparison, and chaining two comparisons. Verifying these three properties makes "same up to isomorphism" a legitimate classification relation. [quotetheorem:9511] The equivalence classes are the objects that classification theorems count. Instead of listing every possible presentation of a group, we choose representatives of the isomorphism classes. [example: Groups of Prime Order] Let $p$ be prime and let $G$ be a group with $|G|=p$. Since $p\ge 2$, the group has an element $g\ne 1_G$. The subgroup $\langle g\rangle$ contains both $1_G$ and $g$, so $|\langle g\rangle|\ge 2$. By *[Lagrange's theorem](/theorems/782)*, $|\langle g\rangle|$ divides $|G|=p$. Since the only positive divisors of the prime $p$ are $1$ and $p$, this forces \begin{align*} |\langle g\rangle|=p. \end{align*} Because $\langle g\rangle\subseteq G$ and both sets have $p$ elements, $\langle g\rangle=G$. Thus $G$ is cyclic, generated by $g$. Now define $\varphi:\mathbb{Z}/p\mathbb{Z}\to G$ by \begin{align*} \varphi(\bar{k})=g^k. \end{align*} This is well-defined: if $\bar{k}=\bar{\ell}$ in $\mathbb{Z}/p\mathbb{Z}$, then $k-\ell=pq$ for some $q\in\mathbb{Z}$, so \begin{align*} g^k=g^{\ell+pq}=g^\ell(g^p)^q=g^\ell 1_G^q=g^\ell. \end{align*} For $\bar{a},\bar{b}\in\mathbb{Z}/p\mathbb{Z}$, addition of residue classes gives $\bar{a}+\bar{b}=\overline{a+b}$, and hence \begin{align*} \varphi(\bar{a}+\bar{b})=\varphi(\overline{a+b})=g^{a+b}=g^ag^b=\varphi(\bar{a})\varphi(\bar{b}). \end{align*} So $\varphi$ is a group homomorphism. The elements \begin{align*} 1_G,\ g,\ g^2,\ \ldots,\ g^{p-1} \end{align*} are pairwise distinct. Indeed, if $g^r=g^s$ with $0\le r<s\le p-1$, then multiplying by $g^{-r}$ gives \begin{align*} g^{s-r}=1_G, \end{align*} where $1\le s-r\le p-1$, contradicting that $g$ has order $p$. Since $G=\langle g\rangle$, these $p$ elements are all of $G$. Therefore $\varphi$ is both injective and surjective, so it is an isomorphism. Hence every group of order $p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, and there is exactly one isomorphism class of groups of order $p$. [/example] This example captures the classification meaning of isomorphism. There may be many ways to present or realize a group of prime order, but they all have the same group structure. ## Beyond and Connected Topics Group isomorphism is the entry point to classification in algebra. The group theory developed in Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups), uses isomorphism to treat groups as algebraic structures rather than as particular sets with chosen names for their elements. The next major step is the [first isomorphism theorem](/theorems/791). A homomorphism $\varphi: G \to H$ that is not itself an isomorphism still produces one after quotienting by its kernel: $G/\ker\varphi$ is isomorphic to $\operatorname{im}\varphi$. This idea is part of the broader algebraic language developed in Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Automorphism groups turn isomorphism inward. They connect naturally to group actions, conjugacy, semidirect products, and Galois groups, because symmetries of mathematical objects often form groups under composition. The same philosophy appears throughout algebra. [Vector space](/page/Vector%20Space) isomorphisms preserve linear structure, module isomorphisms preserve addition and scalar multiplication, and ring isomorphisms preserve addition and multiplication. Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), gives the linear version of this viewpoint, while Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra), develops the ring and module versions in a more advanced setting. ## References Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Dummit and Foote, *Abstract Algebra* (2004). Gallian, *Contemporary Abstract Algebra* (2021). Rotman, *An Introduction to the Theory of Groups* (1995).