Monoid homomorphism and group homomorphism

Let $(M,\cdot ,e)$ and $(M^{\mathrm{\prime }},{\cdot }^{\mathrm{\prime }},e^{\mathrm{\prime }})$ be monoids (Semigroup and monoid). A monoid homomorphism from $M$ to $M^{\mathrm{\prime }}$ is a function $f:M\to M^{\mathrm{\prime }}$ such that

• (H1) $f(x\cdot y)=f(x){\cdot }^{\mathrm{\prime }}f(y)$ for all $x,y\in M$;
• (H2) $f(e)=e^{\mathrm{\prime }}$.

Let $G$ and $G^{\mathrm{\prime }}$ be groups (Group and abelian group). A group homomorphism from $G$ to $G^{\mathrm{\prime }}$ is a function $f:G\to G^{\mathrm{\prime }}$ satisfying (H1) alone:

$f(xy)=f(x)f(y)\text{for all }x,y\in G\mathrm{.}$

Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(x^{-1})=f(x{)}^{-1}$ (A group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(g^{-1})=f(g{)}^{-1}$, and $f(g^n)=f(g{)}^n$ for every $n\in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed). For monoids it does not follow and must be assumed, which is why the two definitions differ.

A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g\circ f)(xy)=g(f(x)f(y))=g(f(x))g(f(y))$ and $(g\circ f)(e)=g(e^{\mathrm{\prime }})=e^{\mathrm{\prime }\mathrm{\prime }}$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism.

• (H1) is a statement about two different operations. On the left the product is formed in $M$, on the right in $M^{\mathrm{\prime }}$; the notation suppresses that and the reader must supply it. The definition says exactly that $f$ turns products into products, and nothing else.

• The asymmetry between the two definitions is real, not stylistic. The map $\mathbb{Z}\to \mathbb{Z}$ sending every integer to $0$ satisfies (H1) for the multiplicative monoid of $\mathbb{Z}$, since $0=0\cdot 0$, and it sends the identity $1$ to $0\mathrm{\ne }1$, so it is not a monoid homomorphism. In a group the same phenomenon is impossible, and cancellation is the reason (A group homomorphism automatically satisfies $f(e)=e^{\mathrm{\prime }}$ and $f(g^{-1})=f(g{)}^{-1}$, and $f(g^n)=f(g{)}^n$ for every $n\in \mathbb{Z}$; for monoid homomorphisms preservation of the identity must be assumed).

• Only the definition is given here. Kernels, images, isomorphisms and the isomorphism theorems belong to a later page; nothing on this page or its companion uses them.