Monoid homomorphism and group homomorphism

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

• (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$;
• (H2) $f(e) = e'$.

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

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

Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ and $f(x^{-1}) = f(x)^{-1}$ (A group homomorphism automatically satisfies $f(e) = e'$ 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 $(g \circ f)(e) = g(e') = e''$; 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 $M'$; 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 $0 \ne 1$, 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 $f(e) = e'$ 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.