Projection on Group Direct Product is Epimorphism/Proof 1
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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
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:
\(\ds \map {\pr_1} {g \circ h}\) | \(=\) | \(\ds \map {\pr_1} {\tuple {g_1, g_2} \circ \tuple {h_1, h_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\pr_1} {\tuple {g_1 \circ_1 h_1, g_2 \circ_2 h_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g_1 \circ_1 h_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\pr_1} g \circ_1 \map {\pr_1} h\) |
\(\ds \map {\pr_2} {g \circ h}\) | \(=\) | \(\ds \map {\pr_2} {\tuple {g_1, g_2} \circ \tuple {h_1, h_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\pr_2} {\tuple {g_1 \circ_1 h_1, g_2 \circ_2 h_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g_2 \circ_2 h_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\pr_2} g \circ_2 \map {\pr_2} h\) |
and thus the morphism property is demonstrated for both $\pr_1$ and $\pr_2$.
$\blacksquare$