Direct Product of Group Homomorphisms is Homomorphism
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Theorem
Let $\struct {G, \circ}, \struct {H_1, *_1}$ and $\struct {H_2, *_2}$ be groups.
Let $\struct {H_1 \times H_2, *}$ be the group direct product of $H_1$ and $H_2$.
Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be group homomorphisms.
Let $f_1 \times f_2: g \to H_1 \times H_2$ be the direct product of $f_1$ and $f_2$.
Then $f_1 \times f_2$ is a group homomorphism.
Proof
The direct product of $f_1$ and $f_2$ $f_1 \times f_2: g \to H_1 \times H_2$ is defined as:
- $\forall g \in G: \map {\paren {f_1 \times f_2} } g = \tuple {\map {f_1} g, \map {f_2} g}$
From External Direct Product of Groups is Group, the group direct product $H_1 \times H_2$ is a group.
It remains to be shown that $f_1 \times f_2$ fulfils the morphism property.
Let $g, h \in G$.
Then:
\(\ds \) | \(\) | \(\ds \map {\paren {f_1 \times f_2} } g * \map {\paren {f_1 \times f_2} } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\map {f_1} g, \map {f_2} g} * \tuple {\map {f_1} h, \map {f_2} h}\) | Definition of Direct Product of $f_1$ and $f_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\map {f_1} g *_1 \map {f_1} h, \map {f_2} g *_2 \map {f_2} h}\) | Definition of Group Direct Product $\struct {H_1 \times H_2, *}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\map {f_1} {g \circ h}, \map {f_2} {g \circ h} }\) | $f_1$ and $f_2$ are Group Homomorphisms | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f_1 \times f_2} } {g \circ h}\) | Definition of Direct Product of $f_1$ and $f_2$ |
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Direct Products