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 $N_0,\ldots,N_{r-1}$ be normal subgroups, where $r\in\mathbb N$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}.$ 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 $(g,h)^{-1}=(g^{-1},h^{-1}).$ 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 $(M',\cdot',e')$ be monoids (def-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 (def-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 $f(x^{-1}) = f(x)^{-1}$ (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 $(g \circ f)(e) = g(e') = e''$; 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 $N\mathrel{\trianglelefteq}G$, then $HN$ is a subgroup and $H\cap N\mathrel{\trianglelefteq}H$. Here $HN:=\{hn:h\in H,\ n\in N\}$. (If $H\le G$ and $N\mathrel{\trianglelefteq}G$, then $HN$ is a subgroup and $H\cap N\mathrel{\trianglelefteq}H$).