# The Canonical Injections of Weak Direct Products of Groups

Recall from The Weak Direct Product of an Arbitrary Collection of Groups page that if $\{ G_i : i \in I \}$ is an arbitrary collection of groups the the weak direct product of the groups $\{ G_i : i \in I \}$ is the set:

(1)with the operation of pointwise product, defined for all $f, g \in \prod_{i \in I}^{\mathrm{weak}} G_i$ by $(fg)(i) = f(i)g(i)$ for all $i \in I$. We proved that $\prod_{i \in I}^{\mathrm{weak}} G_i$ is a normal subgroup of $\prod_{i \in I} G_i$.

We will now define some important functions related to the weak direct product of groups.

Definition: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. The Canonical Injection of $G_j$ into $\prod_{i \in I}^{\mathrm{weak}}$ is the map $\iota_j : G_j \to \prod_{i \in I}^{\mathrm{weak}}$ defined for all $g \in G$ by $\iota_j(g)$ to be the function defined for all $i \in I$ by $[\iota_j(g)](i) = \begin{Bmatrix} g & \mathrm{if} \: i = j \\ e_{G_i} & \mathrm{if} \: i \neq j \end{Bmatrix}$. |

*In other words, each $\iota_j(g)$ is the function on $I$ that maps every $i \in I$ to $e_{G_i}$ with the exception that $\iota_j(g)$ maps $j$ to $g$.*

The following proposition tells us that each canonical injection $\iota_j$ is a monomorphism from $G_j$ to $\prod_{i \in I}^{\mathrm{weak}} G_i$.

Proposition 1: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. Then for each $j \in I$, the canonical injection $\iota_j : G_j \to \prod_{i \in I}^{\mathrm{weak}} G_i$ is a monomorphism. |

**Proof:**Let $j \in I$ and let $g, g' \in G_J$. Then:

- Therefore:

- So indeed, for all $g, g' \in G_j$ we have that $\iota_j(gg') = \iota_j(g) \iota_j(g)$, so $\iota_j$ is a homomorphism.

- Now let $g, g' \in G_j$ and suppose that $\iota_j(g) = \iota_j(g')$. Then $[\iota_j(g)](j) = [\iota_j(g')](j)$, or equivalently, $g = g' 4]]. So [[$ \iota_j$ is injective.

- Thus $\iota_j : G_j \to \prod_{i \in I}^{\mathrm{weak}} G_i$ is a monomorphism. $\blacksquare$

Proposition 2: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. Then for each $j \in J$, $\iota_j(G_j)$ is a normal subgroup of $\prod_{i \in I} G_i$. |

**Proof:**By Proposition 1, $\iota_j$ is a homomorphism of $G_j$ to $\prod_{i \in I}^{\mathrm{weak}} G_i$ and so $\iota_j(G_j)$ is a subgroup of $\prod_{i \in I}^{\mathrm{weak}} G_i$. Furthermore, $\prod_{i \in I}^{\mathrm{weak}} G_i$ is a subgroup of $\prod_{i \in I} G_i$. So $\iota_j(G_j)$ is a subgroup of $\prod_{i \in I} G_i$.

- Let $G = \prod_{i \in I} G_i$ and let $H = \iota_j(G_j)$. We aim to show that for all $g \in G$ that $gHg^{-1} \subseteq H$.

- Let $g \in G$ and let $h \in H$. Since $h \in H$ there exists an $a \in G_j$ such that $\iota_j(a) = h$. So $h(j) = a$ and $h(i) = e_{G_i}$ for all $i \in I \setminus \{ j \}$.

- So if $i = j$ we have that:

- And if $i \in I \setminus \{ j \}$ we have that:

- Let $ghg^{-1} = \iota_j(g(j)ag^{-1}(j)) \in \iota_j(G_j)$. Thus $gHg^{-1} \subseteq H$ which shows that $H = \iota_j(G_j)$ is a normal subgroup of $\prod_{i \in I} G_i$. $\blacksquare$