It is known that the homomorphisms between group preserve products. However, the anti-homomorphisms between groups invert products in contrast to the homomorphisms. In this article, we briefly introduce the notion of anti-homomorphisms of groups, and we will study anti-homomorphisms of rings in the next article.1. Anti-homomorphisms of groups.In this section, all groups are not necessarily abelian. For simplicity, we will simply call anti-homomorphism instead of anti-homomorphism of groups.Definition 1. An anti-homomorphism $\phi:(A,\cdot) \rightarrow (B,\cdot)$ is a mapping $\phi : A \rightarrow B$ such that $\phi(x \cdot y) = \phi(y) \cdot \phi(x)$ for all $x, y \in A$.Anti-homomorphism preserves identity elements, inverses, and powers, which can be readily proved.Proposition 2. If $\phi: A \rightarrow B$ is an anti-homomorphism of groups, then $\phi(1) = 1$, $\phi(x^{−1}) = \phi(x)^{−1}$, and $\phi(x^{n}) = \phi(x)^{n}$ , for all $x\in A, n \in \mathbb{Z}$.Compared to monomorphisms, epimorphisms, and isomorphisms of groups, we could define anti-monomorphisms, anti-epimorphisms and anti-isomorphisms. Before that, first consider the following proposition:Proposition 3. If $\varphi$ is a bijective anti-homomorphism of groups, then its inverse bijection $\varphi^{−1}$ is also an anti-homomorphism of groups.Definition 4. An anti-monomorphism of groups is an injective anti-homomorphism of groups. An anti-epimorphism of groups is a surjective anti-homomorphism of groups. An anti-isomorphism of groups is a bijective anti-homomorphism of groups. Two groups $A$ and $B$ are called anti-isomorphic when there exists an anti-isomorphism $A\cong B$. If $f$ is an anti-isomorphism, then its inverse $f^{−1}$ is called its anti-inverse.Next, we define kernels and images of anti-homomorphisms.Definition 5. Let $\varphi : A\rightarrow B$ be an anti-homomorphism of groups. Then the kernel of $\phi$ is $$\textrm{Ker }\varphi:=\{a\in A\mid \varphi(a)=1\}.$$The image of $\phi$ is$$\textrm{Im }\varphi:=\{\varphi(a)\mid a\in A\}.$$We note some propositions about kernel, cokernel, and image of an anti-homomorphism of groups.Proposition 6. Let $\varphi:A\rightarrow B$ be an anti-homomorphism of groups. Then ${\rm{Ker}}\varphi$ is a normal subgroup of $A$ and ${\rm{Im}}\varphi$ is a subgroup of $B$.Proof. We just prove that $\textrm{Ker }\varphi$ as a subgroup of $A$ is normal. Since in Anti-homomorphism in Rings, we have $\varphi(x)=\varphi(y)$ if and only if $x-y\in\textrm{Ker }\varphi$ for any $x,y\in A$. Then $x\in\textrm{Ker }\varphi+y$ if and only if $x\in y+\textrm{Ker }\varphi$, which shows that $\textrm{Ker }\varphi$ is normal in $A$.Proposition 7. Let $\varphi:A\rightarrow B$ be an anti-homomorphism of groups. Then $\varphi$ is an anti-monomorphism if and only if ${\rm{Ker }}\varphi=0$. And $\varphi$ is an anti-epimorphism if and only if ${\rm{Coker}} \varphi=0$.

ABSTRACTIn this chapter, the concept of anti-homomorphisms in Rings is studied and many results are established.INTRODUCTIONThe concept of anti–homomorphism in groups and rings is not found much in the literature. Authors like Jocobson Neal McCoy and Zariski and Samul have touched this concept in a lighter way. The reason is perhaps that the composition of two anti-homomorphisms is not an anti-homomorphism, but a homomorphism. This discouraged the mathematicians to move further. In 2006, G. Gopalakrishnamoorthy defined a new composition of two anti-homomorphisms so that the composition of two anti-homomorphisms with respect to the new composition is again an anti-homomorphism. In this paper we study the concept of anti-homomorphisms in Rings. Definition. Let $R1$ and $R2$ be two Rings (not necessarily commutative). A map $f: R1\rightarrow R2$ is said to be an anti-homomorphism if i) $f(x+y) = f(x) +f(y)$ and ii) $f(xy) = f(y)\cdot f(x)$ for all $x, y\in R1$. Definition. Let $f: R1\rightarrow R2$ be an anti-homomorphism of Rings. (a) If $f$ is one-one, we say $f$ is an anti-monomorphism. (b) If $f$ is onto, we say $f$ is an anti-epimorphism. (c) If $f$ is both one-one and onto, we say $f$ is an anti-isomorphism. (d) If $R1=R2$ and $f$ is onto, we say $f$ is an anti-endomorphism on $R1$. (e) If $R1 = R2$ and $f$ is both one-one and onto, we say $f$ is an anti-automorphism on $R1$. Property 1: Let $f : X \rightarrow Y$ be a map. Let $A1, A2$ be subsets of $X$. Then (i) $f (\varphi) = \varphi$ (ii) $A1 \subset A2 => f (A1) \subset f(A2) $(iii) $f(A1\cup A2) = f(A1) \cup f(A2) $(iv) $f(A1\cap A2) \subset f(A1) \cap f(A2) $(v) $f(A1\ A2) \supseteq f(A1) \ f(A2)$Property 2: Let $f : X \rightarrow Y$ be a map. Let $B1, B2$ be subsets of $Y$. Then (i) $f^{-1} (\phi)=\phi$ (ii) $B1 \subset B2 \Rightarrow f^{ -1} (B1) \subset f^{ -1} (B2) $(iii) $f^{ -1} (B1\cup B2) = f ^{ -1} (B1)\cup f^{ -1} (B2) $(iv) $f^{ -1} (B1\cap B2) = f ^{ -1} (B1)\cap f^{ -1} (B2) $(v) $f^{ -1} (B1 \ B2) = f ^{ -1} (B1) -f ^{ -1} (B2) $(vi) $f^{ -1} (Y- B1) = X -f^{ -1} (B1)$ (vii) $f^{ -1} (B1) = \phi \Rightarrow B1 \cap f(x) = \phi $Theorem : Let $f : R1\rightarrow R2$ be an anti-homomophism of rings. Then the following hold(i) $f(0) = 0$ (ii) $f(-a) = - f(a)$ (iii) $f(a^{k}) = [f(a)]^{k}$ for all $k \geq 1$, and $a \in R1$Proof: (i) $f(0) = f(0+0) = f(0) + f(0) \Rightarrow f(0) = 0 $(ii) $0 = f(0) = f(a + (-a)) 0 = f(a) + f(-a) \Rightarrow f(-a) = - f(a)$ (iii) $f(a^{2}) = f(a)\cdot f(a) = [f(a^{2})] $Assume $f(a^{k-1} ) = [f(a)]^{k-1}$$$\begin{align} f(a^{k} ) &= f(a^{k-1} \cdot a) \\ &= f(a) \cdot f(a^{k-1}) \\ &= f(a) \cdot [f(a)]^{k-1} \\ &= [f(a)]^{k} \end{align}$$. Theorem : An anti-homomorphism $f : R1 \rightarrow R2$ is a homomorphism if and only if $f(R1)$ is a commutative subring of $R2$.Proof: Let $f : R1 \rightarrow R2$ be an anti-homomorphism of rings. Assume $f(R1)$ is commutative subring of $R2$. Then for all $x,y \in R1 f(xy) = f(y)\cdot f(x) = f(x)\cdot f(y) $. Thus, $f(R1)$ is a commutative subring of $R2$ such that $f$ is a homomorphism. Conversely, assume that $f$ is a homomorphism. Let $a,b \in f(R1)$ be any two elements. Then $a = f(x)$ for some $x\in R1 ,b = f(y)$ for some $y\in R1$. Now $$\begin{align} ab &= f(x) \cdot f(y) \\ &= f(xy) (\textrm{so f is a homomorphism}) \\ &= f(y) . f(x) (\textrm{so f is an anti-homomorphism}) \\ &= b . a \end{align} $$Hence $f(R1)$ is a commutative subring of R2. Theorem : Let $f : R1 \rightarrow R2$ be an anti-homomorphism of rings. Then (i) $\textrm{Ker}f$ is a subring of $R1$. (ii) $f(R1)$ is a subring of $R2$. Proof: (i) Since $R1$ is non-empty, $f(R1)$ is non-empty.Let $a,b \in f (R1)$. Then a = f(x) for some $x\in R1, b = f(y)$ , and for some $y\in R1$.Then $$\begin{align} a-b &= f(x) – f(y)\\ &= f(x) + f(-y) \\ &= f(x-y) \in f(R1)  \end{align}$$Also $$\begin{align} ab &= f(x)\cdot f(y) \\ &= f(yx) \in f(R1) \end{align}$$Thus $f(R1)$ is a subring of $R2$. Theorem : Let $f : R1 \rightarrow R2$ be an anti-homomorphism of rings. For all $x, y \in R1 , f(x) = f(y)$ iff $x-y \in \textrm{ker }f$. Theorem : Let $f : R1\rightarrow R2$ be an anti-homomophism of rings. Then $f$ is one-one iff $\textrm{ker }f = {0}$.