The Canonical Projections of Direct Products of Groups

Recall from The Direct Product of an Arbitrary Collection of Groups page that if $\{ G_i : i \in I \}$ is an arbitrary collection of groups then the direct product of these groups is the set $\prod_{i \in I} G_i$ with the operation defined for all $f, g \in \prod_{i \in I} G_i$ as componentwise product, i.e., for each $f, g \in \prod_{i \in I} G_i$, $fg$ is defined for each $i \in I$ by:

(1)

\begin{align} \quad (fg)(i) = f(i)g(i) \end{align}

We will now define some important maps associated with direct products of groups.

Definition: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. The Canonical Projection Onto the Group $G_j$ is the map $\pi_j : \prod_{i \in I} G_i \to G_j$ defined for all $f \in \prod_{i \in I} G_i$ by $\pi_j(f) = f(j)$.

The following proposition tells us that each $p_j$ is an epimorphism of $\prod_{i \in I} G_i$ onto $G_j$.

Proposition 1: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. Then for each $j \in I$, the canonical projection $\pi_j : \prod_{i \in I} G_i \to G_j$ is an epimorphism.

• Proof: Clearly each $\pi_j$ is an epimorphism since for all $f, g \in \prod_{i \in I} G_i$ we have that:

(2)

\begin{align} \quad \pi_j(fg) = (fg)(j) = f(j)g(j) = \pi_j(f) \pi_j(g) \end{align}

• Furthermore, each $\pi_j$ is surjective since if $g \in G_j$ let $f \in \prod_{i \in I} G_i$ be defined with $f(i) = e_{G_i}$ if $i \in I \setminus \{ j \}$ (where $e_{G_i}$ denotes the identity of $G_i$) and let $f(j) = g$ (of course, any appropriate function $f$ with $f(j) = g$ will do). Then:

(3)

\begin{align} \quad \pi_j(f) = f(j) = g \end{align}

• So $\pi_j$ is onto, i.e., $\pi_j$ is an epimorphism. $\blacksquare$