Internal direct products are external direct products, equivalently every element has a unique factorisation

Given: The objects and hypotheses in the statement.

Let $G$ be a group and let $N_0,\dots ,N_{r-1}$ be normal subgroups, where $r\in \mathbb{N}$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap ⟨N_j:j<r,j\mathrm{\ne }i⟩=\{e\}\mathrm{.}$ The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says $G=HK$ and $H\cap K=\{e\}$; in additive notation one writes $G=H\oplus K$. Normal subgroups and generated subgroups are those of def-normal-subgroup and def-generated-subgroup, and the comparison product is def-external-direct-product-of-groups. (Internal direct products of finitely many normal subgroups).

For groups $G$ and $H$, the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$, and $(g,h{)}^{-1}=(g^{-1},h^{-1})\mathrm{.}$ Moreover the coordinate maps ${\pi }_G(g,h)=g$ and ${\pi }_H(g,h)=h$ are group homomorphisms. ($G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections).

Let $(M,\cdot ,e)$ and $(M^{\mathrm{\prime }},{\cdot }^{\mathrm{\prime }},e^{\mathrm{\prime }})$ be monoids (def-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 (def-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 $f(x^{-1})=f(x{)}^{-1}$ (lem-group-homomorphism-basic-properties). 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. (Monoid homomorphism and group homomorphism).

A group homomorphism is injective if and only if its kernel is trivial. For a group homomorphism $f:G\to H$, $f$ is injective exactly when $\ker f=\{e_G\}$. (A group homomorphism is injective if and only if its kernel is trivial).

If $H\le G$ and $N⊴G$, then $HN$ is a subgroup and $H\cap N⊴H$. Here $HN:=\{hn:h\in H,n\in N\}$. (If $H\le G$ and $N⊴G$, then $HN$ is a subgroup and $H\cap N⊴H$).