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:

\(\displaystyle \) \(\) \(\displaystyle \map {\paren {f_1 \times f_2} } g * \map {\paren {f_1 \times f_2} } h\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {\map {f_1} g, \map {f_2} g} * \tuple {\map {f_1} h, \map {f_2} h}\) Definition of Direct Product of $f_1$ and $f_2$
\(\displaystyle \) \(=\) \(\displaystyle \tuple {\map {f_1} g *_1 \map {f_1} h, \map {f_2} g *_2 \map {f_2} h}\) Definition of Group Direct Product $\struct {H_1 \times H_2, *}$
\(\displaystyle \) \(=\) \(\displaystyle \tuple {\map {f_1} {g \circ h}, \map {f_2} {g \circ h} }\) $f_1$ and $f_2$ are Group Homomorphisms
\(\displaystyle \) \(=\) \(\displaystyle \map {\paren {f_1 \times f_2} } {g \circ h}\) Definition of Direct Product of $f_1$ and $f_2$

$\blacksquare$


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