Projection on Group Direct Product is Epimorphism

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Theorem

Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups.

Let $\struct {G, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.


Then:

$\pr_1$ is an epimorphism from $\struct {G, \circ}$ to $\struct {G_1, \circ_1}$
$\pr_2$ is an epimorphism from $\struct {G, \circ}$ to $\struct {G_2, \circ_2}$

where $\pr_1$ and $\pr_2$ are the first and second projection respectively of $\struct {G, \circ}$.


Proof 1

From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections.

We now need to show they are homomorphisms.

Let $g, h \in \struct {G, \circ}$ where $g = \tuple {g_1, g_2}$ and $h = \tuple {h_1, h_2}$.

Then:

\(\displaystyle \map {\pr_1} {g \circ h}\) \(=\) \(\displaystyle \map {\pr_1} {\tuple {g_1, g_2} \circ \tuple {h_1, h_2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\pr_1} {\tuple {g_1 \circ_1 h_1, g_2 \circ_2 h_2} }\)
\(\displaystyle \) \(=\) \(\displaystyle g_1 \circ_1 h_1\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\pr_1} g \circ_1 \map {\pr_1} h\)


\(\displaystyle \map {\pr_2} {g \circ h}\) \(=\) \(\displaystyle \map {\pr_2} {\tuple {g_1, g_2} \circ \tuple {h_1, h_2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\pr_2} {\tuple {g_1 \circ_1 h_1, g_2 \circ_2 h_2} }\)
\(\displaystyle \) \(=\) \(\displaystyle g_2 \circ_2 h_2\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\pr_2} g \circ_2 \map {\pr_2} h\)

and thus the morphism property is demonstrated for both $\pr_1$ and $\pr_2$.

$\blacksquare$


Proof 2

A specific instance of Projection is Epimorphism, where the algebraic structures in question are groups.

$\blacksquare$


Sources