A group remembers how its elements multiply, but mathematics rarely studies a group in isolation. We compare a symmetry group with a permutation group, reduce an infinite group modulo a [normal subgroup](/page/Normal%20Subgroup), or represent abstract elements as matrices, and the basic question is: what kind of function between groups respects the multiplication enough that group-theoretic information survives the trip? A bijection of underlying sets is too weak. It may rename the elements while destroying products. A map that is not bijective may still be valuable, because it can forget exactly the information we want to discard. The useful notion is not sameness of sets, but compatibility with the group operation. [example: Parity as a Multiplicative Shadow] Let $G=(\mathbb{Z},+)$ and let $H=\{1,-1\}$ under multiplication. Define \begin{align*} \varphi:\mathbb{Z}&\to H\\ n&\mapsto(-1)^n. \end{align*} This is a well-defined function into $H$, because $(-1)^n=1$ when $n$ is even and $(-1)^n=-1$ when $n$ is odd. For all $m,n\in\mathbb{Z}$, the integer exponent law gives \begin{align*} \varphi(m+n)&=(-1)^{m+n}\\ &=(-1)^m(-1)^n\\ &=\varphi(m)\varphi(n). \end{align*} Thus the operation in the domain is transported to the operation in the codomain: adding $m$ and $n$ in $\mathbb{Z}$ corresponds exactly to multiplying $(-1)^m$ and $(-1)^n$ in $H$. The map forgets everything about an integer except its parity, and that parity information is precisely what survives as the sign $1$ or $-1$. [/example] This example is the model for the whole chapter. A homomorphism may compress, embed, relabel, or represent a group, but it always does so by respecting products. Kernels, images, quotient groups, and isomorphism theorems are the language for describing what is preserved and what is lost. ## Definition ### Preserving the Operation The product in the domain and the product in the codomain may be written differently, so the definition must compare both group laws. The condition below is the minimum requirement that every multiplication computation in the domain can be transported into the codomain. [definition: Group Homomorphism] Let $(G,\cdot_G)$ and $(H,\cdot_H)$ be groups. A group homomorphism from $G$ to $H$ is a function $\varphi:G\to H$ such that for all $g_1,g_2\in G$, \begin{align*} \varphi(g_1\cdot_G g_2)=\varphi(g_1)\cdot_H\varphi(g_2). \end{align*} [/definition] After the operations are fixed, one usually writes $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. The shorter notation is safe because the domain and codomain determine which product is being used. The definition mentions only products, but a group also has a distinguished identity and inverse operation. At first this leaves a possible gap: a function might preserve multiplication while mishandling those extra pieces of structure. To use homomorphisms as structure-preserving maps, that ambiguity has to be removed. [quotetheorem:4981] The theorem removes a hidden worry in the definition. A homomorphism is allowed to mention only products because the identity and inverse operations are then forced to behave correctly. This is what lets group homomorphisms preserve equations, powers, orders, and relations without separately checking every piece of the group structure. ### Equations and Relations A homomorphism does more than preserve individual products. It sends every equation built from multiplication and inverses in the domain to a corresponding equation in the codomain. This is why homomorphisms are useful for studying groups given by relations. [example: Relations Survive Under Homomorphisms] Let $\varphi:G\to H$ be a group homomorphism, and suppose $g\in G$ satisfies $g^6=1_G$. We compute the sixth power of $\varphi(g)$ by repeatedly using the homomorphism property: \begin{align*} \varphi(g)^2&=\varphi(g)\varphi(g)=\varphi(g^2),\\ \varphi(g)^3&=\varphi(g)^2\varphi(g)=\varphi(g^2)\varphi(g)=\varphi(g^3),\\ \varphi(g)^4&=\varphi(g)^3\varphi(g)=\varphi(g^3)\varphi(g)=\varphi(g^4),\\ \varphi(g)^5&=\varphi(g)^4\varphi(g)=\varphi(g^4)\varphi(g)=\varphi(g^5),\\ \varphi(g)^6&=\varphi(g)^5\varphi(g)=\varphi(g^5)\varphi(g)=\varphi(g^6). \end{align*} Since $g^6=1_G$, this gives \begin{align*} \varphi(g)^6=\varphi(g^6)=\varphi(1_G)=1_H, \end{align*} where the last equality uses *[Homomorphisms Preserve Identity and Inverses](/theorems/4981)*. Thus $\varphi(g)$ satisfies $x^6=1_H$, so the order of $\varphi(g)$ divides $6$. A homomorphism may reduce the order of an element, but it cannot send a solution of $x^6=1_G$ to an element that violates the corresponding equation in $H$. [/example] This relation-preserving property is the practical reason homomorphisms appear everywhere in algebra. They allow us to replace a group by a more concrete or smaller one while keeping control of all multiplicative identities. ## Kernels and Images The first diagnostic for a homomorphism is to ask which elements become indistinguishable from the identity. Those elements are precisely the information lost by the map, and collecting them turns a question about a function into a question about a subgroup. [definition: Kernel of a Group Homomorphism] 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] The kernel of the parity homomorphism is $2\mathbb{Z}$. More generally, the kernel records the identity fibre of the homomorphism, and all other fibres are its cosets. To use kernels in quotient constructions, we need them to be normal subgroups rather than arbitrary subgroups. [quotetheorem:788] This theorem explains why kernels are the right subgroups for quotient constructions. They are not merely subsets of elements that disappear; they are normal subgroups, so their cosets form a [quotient group](/theorems/790). That is why every homomorphism naturally produces a quotient of its domain. The codomain may contain elements that are never hit by the homomorphism. If we want to separate the group that the map actually reaches from the larger ambient codomain, we need a precise name for that reached part. [definition: Image of a Group Homomorphism] Let $\varphi:G\to H$ be a group homomorphism. The image of $\varphi$ is \begin{align*} \operatorname{im}\varphi=\{h\in H:h=\varphi(g)\text{ for some }g\in G\}. \end{align*} [/definition] When a homomorphism is not surjective onto $H$, replacing $H$ by $\operatorname{im}\varphi$ should isolate the part of the codomain that the domain really sees. To use the image as a genuine target group in later arguments, we must know that the reached elements inherit the group structure from $H$. The needed test is closure under multiplication and inverses inside the image, not merely the fact that the elements are outputs of $\varphi$. [quotetheorem:769] The image theorem lets us replace the ambient codomain by the part that is actually reached. This matters whenever a homomorphism is not surjective: the map may miss most of $H$, but it is always surjective onto $\operatorname{im}\varphi$, and that reached part is itself a group. For injectivity, the problem is not whether the whole image is large, but whether anything besides the identity has been collapsed to the identity. The next criterion turns that idea into a precise test. [quotetheorem:792] The kernel criterion turns injectivity into a structural question about what the homomorphism forgets. If the kernel is trivial, no nonidentity element has been collapsed to the identity, so the map loses no group-theoretic information. This criterion is often the cleanest route to proving that a homomorphism embeds one group inside another. [example: The Sign Homomorphism] Let $S_n$ be the symmetric group on $\{1,\dots,n\}$. Using the standard parity theorem for permutations, define $\operatorname{sgn}(\pi)=1$ when $\pi$ is a product of an even number of transpositions and $\operatorname{sgn}(\pi)=-1$ when $\pi$ is a product of an odd number of transpositions. Let $\sigma,\tau\in S_n$. If \begin{align*} \sigma&=t_1t_2\cdots t_r,\\ \tau&=u_1u_2\cdots u_s \end{align*} are decompositions into transpositions, then \begin{align*} \sigma\tau &=(t_1t_2\cdots t_r)(u_1u_2\cdots u_s), \end{align*} so $\sigma\tau$ is written as a product of $r+s$ transpositions. Therefore \begin{align*} \operatorname{sgn}(\sigma\tau) &=(-1)^{r+s}\\ &=(-1)^r(-1)^s\\ &=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau). \end{align*} Thus $\operatorname{sgn}:S_n\to\{1,-1\}$ is a group homomorphism. Its kernel consists exactly of the permutations with sign $1$: \begin{align*} \ker(\operatorname{sgn}) &=\{\sigma\in S_n:\operatorname{sgn}(\sigma)=1\}\\ &=A_n. \end{align*} For $n\ge 2$, the identity permutation has sign $1$, while the transposition $(1\ 2)$ has sign $-1$. Hence \begin{align*} \operatorname{im}(\operatorname{sgn})=\{1,-1\}. \end{align*} The [sign homomorphism](/theorems/778) collapses all even permutations to $1$ and all odd permutations to $-1$, so it records precisely the parity of a permutation. [/example] The example shows how normal subgroups often arise in practice. Instead of discovering $A_n$ only by checking conjugation, we see it as the kernel of a natural map. ## Quotients and Factorization A homomorphism with nontrivial kernel identifies many elements. The quotient by the kernel is the group obtained by making exactly those identifications before applying the map. This is the conceptual bridge from homomorphisms to quotient groups. To define maps out of a quotient, we must know that choosing a different representative of the same coset does not change the output. The following definition isolates the condition that makes the formula on cosets meaningful. [definition: Homomorphism Induced on a Quotient] Let $G$ and $H$ be groups, let $N\trianglelefteq G$, and let $\varphi:G\to H$ be a group homomorphism such that $N\subset\ker\varphi$. The homomorphism induced by $\varphi$ on $G/N$ is the function \begin{align*} \bar{\varphi}:G/N&\to H\\ gN&\mapsto\varphi(g). \end{align*} [/definition] For a canonical quotient attached to a homomorphism, the subgroup to collapse should be exactly $\ker\varphi$. This choice removes every representative ambiguity and gives an injective comparison between the quotient and the reached subgroup of the codomain. The point of the next theorem is that this is not merely a convenient construction: every homomorphism factors in this precise way, so quotients by kernels are exactly the quotients that homomorphisms naturally create. [quotetheorem:842] The theorem gives the structural anatomy of every homomorphism. The kernel is the part of the domain that becomes invisible, the quotient removes exactly that invisible part, and the image is the part of the codomain actually reached. Thus a homomorphism is best understood as three operations in sequence: collapse the kernel, identify the resulting quotient with the image, and then view that image inside the codomain. [example: A Failed Induced Map] Let $G=\mathbb{Z}$, let $N=3\mathbb{Z}$, and define \begin{align*} \varphi:\mathbb{Z}&\to\mathbb{Z}/2\mathbb{Z}\\ n&\mapsto\bar{n}. \end{align*} The identity element of $\mathbb{Z}/2\mathbb{Z}$ is $\bar{0}$, so \begin{align*} \ker\varphi &=\{n\in\mathbb{Z}:\varphi(n)=\bar{0}\}\\ &=\{n\in\mathbb{Z}:\bar{n}=\bar{0}\}\\ &=2\mathbb{Z}. \end{align*} Now $3\in 3\mathbb{Z}=N$, because $3=3\cdot 1$, but \begin{align*} \varphi(3)=\bar{3}=\bar{1}\ne\bar{0} \end{align*} in $\mathbb{Z}/2\mathbb{Z}$. Hence $3\notin\ker\varphi$, so $N\not\subset\ker\varphi$. The ambiguity appears at the level of representatives. In $\mathbb{Z}/3\mathbb{Z}$, \begin{align*} 0+3\mathbb{Z}=3+3\mathbb{Z}, \end{align*} because $3-0=3\in 3\mathbb{Z}$. If we tried to define a quotient map by \begin{align*} n+3\mathbb{Z}\mapsto\bar{n}, \end{align*} then the representative $0$ would give \begin{align*} 0+3\mathbb{Z}\mapsto\bar{0}, \end{align*} while the representative $3$ of the same coset would give \begin{align*} 3+3\mathbb{Z}\mapsto\bar{3}=\bar{1}. \end{align*} Since $\bar{0}\ne\bar{1}$ in $\mathbb{Z}/2\mathbb{Z}$, the proposed rule is not well-defined. [/example] The example displays exactly why the kernel condition appears in quotient factorization. Choosing the wrong subgroup produces ambiguity before any algebraic structure can be checked. [example: Parity Through the First Isomorphism Theorem] For the homomorphism \begin{align*} \varphi:\mathbb{Z}&\to\{1,-1\}\\ n&\mapsto(-1)^n, \end{align*} we compute its kernel and image. The identity element of $\{1,-1\}$ under multiplication is $1$, so \begin{align*} \ker\varphi &=\{n\in\mathbb{Z}:\varphi(n)=1\}\\ &=\{n\in\mathbb{Z}:(-1)^n=1\}\\ &=\{n\in\mathbb{Z}:n\text{ is even}\}\\ &=2\mathbb{Z}. \end{align*} For the image, every value of $\varphi$ lies in $\{1,-1\}$, and both values occur because \begin{align*} \varphi(0)&=(-1)^0=1,\\ \varphi(1)&=(-1)^1=-1. \end{align*} Hence \begin{align*} \operatorname{im}\varphi=\{1,-1\}. \end{align*} By the *[First Isomorphism Theorem for Groups](/theorems/842)*, the quotient by the kernel is isomorphic to the image: \begin{align*} \mathbb{Z}/\ker\varphi&\cong\operatorname{im}\varphi,\\ \mathbb{Z}/2\mathbb{Z}&\cong\{1,-1\}. \end{align*} Under this isomorphism, the coset $n+2\mathbb{Z}$ is sent to $(-1)^n$, so the quotient keeps exactly whether $n$ is even or odd and discards the rest of the integer. [/example] This theorem is the main computational engine for homomorphisms. To understand a quotient, construct a homomorphism with that kernel; to understand a homomorphism, compute its kernel and image. ## Isomorphisms and Automorphisms ### Reversible Homomorphisms Some homomorphisms preserve all group-theoretic information. A reversible structure-preserving map expresses that two groups differ only in the names or concrete form of their elements. [definition: Group Isomorphism] Let $G$ and $H$ be groups. A group isomorphism from $G$ to $H$ is a bijective group homomorphism $\varphi:G\to H$. [/definition] If there exists an isomorphism $G\to H$, we write $G\cong H$. Isomorphic groups have the same multiplication structure after relabelling. A bijective homomorphism has a set-theoretic inverse, but that alone does not say the inverse respects multiplication. For isomorphism to be a reversible comparison of group structure, the inverse map must also be a homomorphism. The key point to verify is that bijectivity lets every product in the target be pulled back uniquely to a product in the source. This raises the basic test for the definition: once a homomorphism is known to be bijective, does reversibility come for free, or must it be imposed as an extra condition? The following result settles that point and justifies treating isomorphisms as genuinely two-sided structure-preserving maps. [quotetheorem:770] This result makes isomorphism the correct notion of structural sameness for groups. A bijective homomorphism is not just a reversible function whose forward direction respects multiplication; its inverse also respects the group law. Therefore any theorem stated in purely group-theoretic language can be transported across an isomorphism and read in either group. [example: Residue Classes and Roots of Unity] Let $n$ be a positive integer, and let $\mu_n=\{z\in\mathbb{C}:z^n=1\}$ under multiplication. Define \begin{align*} \varphi:\mathbb{Z}/n\mathbb{Z}&\to\mu_n\\ \bar{k}&\mapsto e^{2\pi i k/n}. \end{align*} First, the value lies in $\mu_n$, since \begin{align*} \left(e^{2\pi i k/n}\right)^n &=e^{2\pi i k}\\ &=\left(e^{2\pi i}\right)^k\\ &=1^k\\ &=1. \end{align*} The rule is well-defined: if $\bar{k}=\bar{\ell}$ in $\mathbb{Z}/n\mathbb{Z}$, then $k-\ell=qn$ for some $q\in\mathbb{Z}$, so \begin{align*} e^{2\pi i k/n} &=e^{2\pi i(\ell+qn)/n}\\ &=e^{2\pi i\ell/n}e^{2\pi iq}\\ &=e^{2\pi i\ell/n}\left(e^{2\pi i}\right)^q\\ &=e^{2\pi i\ell/n}. \end{align*} Addition of residue classes is transported to multiplication in $\mu_n$, because for all $\bar{k},\bar{\ell}\in\mathbb{Z}/n\mathbb{Z}$, \begin{align*} \varphi(\bar{k}+\bar{\ell}) &=\varphi(\overline{k+\ell})\\ &=e^{2\pi i(k+\ell)/n}\\ &=e^{2\pi i k/n}e^{2\pi i\ell/n}\\ &=\varphi(\bar{k})\varphi(\bar{\ell}). \end{align*} Thus $\varphi$ is a group homomorphism. It is injective: if $\varphi(\bar{k})=\varphi(\bar{\ell})$, then \begin{align*} e^{2\pi i k/n}&=e^{2\pi i\ell/n},\\ e^{2\pi i(k-\ell)/n}&=1. \end{align*} Since $e^{2\pi i x}=1$ exactly when $x\in\mathbb{Z}$, we get $(k-\ell)/n\in\mathbb{Z}$, so $n\mid(k-\ell)$ and therefore $\bar{k}=\bar{\ell}$. It is surjective: if $z\in\mu_n$, then $z^n=1$, so $|z|^n=1$ and hence $|z|=1$. Write $z=e^{i\theta}$ for some real $\theta$. Then \begin{align*} 1=z^n=(e^{i\theta})^n=e^{in\theta}, \end{align*} so $n\theta=2\pi m$ for some $m\in\mathbb{Z}$. Hence \begin{align*} z=e^{i\theta}=e^{2\pi i m/n}=\varphi(\bar{m}). \end{align*} Therefore $\varphi$ is a bijective homomorphism, so $\mathbb{Z}/n\mathbb{Z}\cong\mu_n$. This identifies the additive cyclic group of residue classes with the multiplicative cyclic group of $n$-th roots of unity. [/example] The example replaces an abstract cyclic group with a concrete multiplicative group in the complex plane. This is one of the most common uses of isomorphisms. ### Symmetries of a Single Group When the domain and codomain are the same group, a reversible homomorphism describes an internal symmetry. Such symmetries are useful because they identify which features of the group are intrinsic. [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 theorem shows that inverse automorphisms remain automorphisms. This motivates collecting all automorphisms into a group of symmetries. [definition: Automorphism Group] Let $G$ be a group. The automorphism group of $G$ is \begin{align*} \operatorname{Aut}(G)=\{\varphi:G\to G:\varphi\text{ is a group automorphism}\} \end{align*} with operation given by composition of functions. [/definition] The definition is useful only if this collection is closed under composition and inverses and has an identity element for composition. Otherwise $\operatorname{Aut}(G)$ would be only a list of symmetries, not a group that can itself be studied algebraically. [quotetheorem:4982] Thus the symmetries of a group become another object of group theory. The group operation in $\operatorname{Aut}(G)$ is composition, so automorphisms can be combined, inverted, and compared internally. This turns questions about the self-symmetry of $G$ into ordinary group-theoretic questions about $\operatorname{Aut}(G)$. ## Generators and Relations Specifying a homomorphism element by element is often impossible. If a group is generated by a smaller set, then the homomorphism is determined by its values on that set, provided the values respect the relations. The notion of generation makes this strategy precise. It identifies the smallest subgroup forced by a chosen collection of elements. [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] If $S$ generates $G$, every element of $G$ is built from elements of $S$ and their inverses. This makes generators a possible shortcut for defining or comparing homomorphisms: instead of checking every element of $G$ separately, we want to know when agreement on $S$ forces agreement everywhere. The obstruction is that different words in the generators can name the same group element, so the theorem isolates the uniqueness part of this strategy. [quotetheorem:4983] This is a uniqueness theorem. Existence still requires the proposed images of the generators to satisfy the relations that hold in $G$. [example: Maps Out of a Cyclic Group] Let $G=\langle g\rangle$ be cyclic of order $n$, and let $H$ be a group. We show that homomorphisms $\varphi:G\to H$ are exactly the maps obtained by choosing an element $h\in H$ with $h^n=1_H$ and sending $g^k$ to $h^k$. First suppose $\varphi:G\to H$ is a homomorphism, and put $h=\varphi(g)$. Since $G=\langle g\rangle$, every element of $G$ has the form $g^k$ for some $k\in\mathbb{Z}$. For $k\ge 1$, repeated use of the homomorphism property gives \begin{align*} \varphi(g^k) &=\varphi(\underbrace{gg\cdots g}_{k\text{ factors}})\\ &=\underbrace{\varphi(g)\varphi(g)\cdots\varphi(g)}_{k\text{ factors}}\\ &=h^k. \end{align*} Also $\varphi(1_G)=1_H$ and $\varphi(g^{-1})=\varphi(g)^{-1}=h^{-1}$ by *[Homomorphisms Preserve Identity](/theorems/768) and Inverses*, so the same formula holds for $k=0$ and for negative $k$. Thus $\varphi$ is determined by the single value $h=\varphi(g)$. Because $g$ has order $n$, we have $g^n=1_G$. Therefore \begin{align*} h^n &=\varphi(g)^n\\ &=\varphi(g^n)\\ &=\varphi(1_G)\\ &=1_H. \end{align*} So the chosen value of $g$ must be an element of $H$ satisfying $h^n=1_H$. Conversely, let $h\in H$ satisfy $h^n=1_H$, and define \begin{align*} \psi:G&\to H\\ g^k&\mapsto h^k. \end{align*} This rule is well-defined. If $g^k=g^\ell$, then \begin{align*} g^{k-\ell}=1_G, \end{align*} so $n\mid(k-\ell)$. Write $k-\ell=qn$ for some $q\in\mathbb{Z}$. Then \begin{align*} h^k &=h^{\ell+qn}\\ &=h^\ell(h^n)^q\\ &=h^\ell 1_H^q\\ &=h^\ell. \end{align*} Thus different exponents representing the same element of $G$ give the same value in $H$. Finally, for any $g^a,g^b\in G$, \begin{align*} \psi(g^ag^b) &=\psi(g^{a+b})\\ &=h^{a+b}\\ &=h^ah^b\\ &=\psi(g^a)\psi(g^b), \end{align*} so $\psi$ is a homomorphism. Hence homomorphisms from a cyclic group of order $n$ correspond exactly to elements $h\in H$ with $h^n=1_H$, or equivalently to elements of $H$ whose order divides $n$. [/example] This is the prototype for presentations: generators may be assigned freely only after all relations have been checked. ## Conjugation and Inner Symmetry Groups often act by symmetries of other groups. The most important internal example is conjugation, where each element of a group gives a symmetry of the whole group. For a fixed element $a\in G$, conjugation by $a$ rearranges $G$ while respecting multiplication. This produces a canonical automorphism associated to $a$. [definition: Inner Automorphism] Let $G$ be a group and let $a\in G$. The inner automorphism induced by $a$ is the function \begin{align*} \iota_a:G&\to G\\ g&\mapsto aga^{-1}. \end{align*} [/definition] A family of automorphisms becomes more useful when it is controlled by a single homomorphism. Conjugation assigns an automorphism to each element of $G$, but to compare $G$ with its automorphism group we need this assignment to preserve the relevant operations. The question is whether multiplying elements of $G$ corresponds to composing their conjugation actions in $\operatorname{Aut}(G)$. [quotetheorem:4984] The kernel of this homomorphism consists of elements whose conjugation action is the identity on every element of $G$. Naming this kernel leads to one of the basic invariants of a group. [definition: Centre of a Group] Let $G$ be a group. The centre of $G$ is \begin{align*} Z(G)=\{z\in G:zg=gz\text{ for all }g\in G\}. \end{align*} [/definition] The quotient by $Z(G)$ should measure exactly how many distinct inner conjugation symmetries the group has. The obstruction is that different elements of $G$ can induce the same conjugation action precisely when they differ by an element that commutes with everything. [quotetheorem:4985] This is a clean example of the general method: construct a homomorphism, identify its kernel, and read off a quotient from its image. [example: Abelian Groups Have No Nonidentity Inner Conjugation] Let $G$ be abelian. For every $a,g\in G$, the inner automorphism induced by $a$ sends $g$ to $aga^{-1}$. Since $G$ is abelian, $ga^{-1}=a^{-1}g$, so \begin{align*} \iota_a(g) &=aga^{-1}\\ &=a(ga^{-1})\\ &=a(a^{-1}g)\\ &=(aa^{-1})g\\ &=1_Gg\\ &=g. \end{align*} Thus $\iota_a(g)=g$ for every $g\in G$, so $\iota_a=\operatorname{id}_G$ for every $a\in G$. Therefore every inner automorphism is the identity automorphism, and since $\operatorname{id}_G=\iota_{1_G}$, we have \begin{align*} \operatorname{Inn}(G)=\{\operatorname{id}_G\}. \end{align*} Also every element commutes with every element of an abelian group, so \begin{align*} Z(G) &=\{z\in G:zg=gz\text{ for all }g\in G\}\\ &=G. \end{align*} Hence the quotient comparison becomes \begin{align*} G/Z(G)=G/G\cong\{\operatorname{id}_G\}, \end{align*} so abelian groups have no nonidentity inner conjugation symmetries. [/example] Non-abelian groups become richer because conjugation supplies nonidentity homomorphisms into automorphism groups. ## Failure Modes A function can look natural and still fail to be a homomorphism. The issue is always the same: the function must preserve the chosen group operation, not some unrelated operation available in the surrounding set. [example: Squaring Need Not Be a Homomorphism] Let $G$ be a group and define \begin{align*} f:G&\to G\\ g&\mapsto g^2. \end{align*} The homomorphism condition for this particular function is that, for every $g,h\in G$, \begin{align*} f(gh)=f(g)f(h). \end{align*} Substituting the definition of $f$ into both sides gives \begin{align*} (gh)^2&=g^2h^2,\\ ghgh&=gghh. \end{align*} We now isolate what this condition says about $g$ and $h$. Starting from \begin{align*} ghgh=gghh, \end{align*} left-multiply both sides by $g^{-1}$: \begin{align*} g^{-1}(ghgh)&=g^{-1}(gghh),\\ (g^{-1}g)hgh&=(g^{-1}g)ghh,\\ 1_Ghgh&=1_Gghh,\\ hgh&=ghh. \end{align*} Then right-multiply both sides by $h^{-1}$: \begin{align*} (hgh)h^{-1}&=(ghh)h^{-1},\\ hg(hh^{-1})&=g(hh^{-1}),\\ hg1_G&=g1_G,\\ hg&=gh. \end{align*} Thus the condition $f(gh)=f(g)f(h)$ for all $g,h\in G$ is equivalent to $hg=gh$ for all $g,h\in G$. If $G$ is abelian, then $gh=hg$ for all $g,h\in G$, and \begin{align*} f(gh)&=(gh)^2\\ &=ghgh\\ &=g(hg)h\\ &=g(gh)h\\ &=g^2h^2\\ &=f(g)f(h), \end{align*} so the squaring map is a homomorphism. If $G$ has elements $g,h$ with $gh\ne hg$, then the calculation above shows that $f(gh)=f(g)f(h)$ cannot hold for that pair. Therefore squaring is a homomorphism exactly on abelian groups, and it fails as a homomorphism whenever the group contains noncommuting elements. [/example] This example is a useful warning: formulas involving powers often behave well in abelian groups and break when order matters. ## Beyond and Connected Topics Group homomorphisms lead directly to [quotient groups](/page/Quotient%20Group), because kernels are normal subgroups and the [First Isomorphism Theorem](/theorems/791) explains why quotients naturally appear. This is the next structural layer after the definition. They also encode [group actions](/page/Group%20Action). An action of $G$ on a set $X$ can be viewed as a homomorphism from $G$ to the group of bijections of $X$, so actions turn abstract group elements into concrete permutations. In linear algebra, group representations are homomorphisms $G\to GL(V)$, where $V$ is a [vector space](/page/Vector%20Space). This connects group theory with [linear maps](/page/Linear%20Map), invariant subspaces, matrices, and characters. In ring theory and module theory, the same philosophy reappears with more structure to preserve. Ring homomorphisms preserve addition, multiplication, and identity; module homomorphisms preserve addition and scalar multiplication. The group case is the clean model for these later maps. In commutative algebra, homomorphisms become geometric through spectra: a ring homomorphism $R\to S$ induces a map $\operatorname{Spec}(S)\to\operatorname{Spec}(R)$ by preimage of prime ideals. This is one reason homomorphisms remain central far beyond elementary algebra. ## 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). Androma, [Quotient Group](/page/Quotient%20Group). Androma, [Group Action](/page/Group%20Action). Androma, [Linear Map](/page/Linear%20Map). Dummit and Foote, *Abstract Algebra* (2004). Lang, *Algebra* (2002). Rotman, *An Introduction to the Theory of Groups* (1995).