Composite of Group Homomorphisms is Homomorphism/Proof 2

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Theorem

Let:

$\struct {G_1, \circ}$
$\struct {G_2, *}$
$\struct {G_3, \oplus}$

be groups.

Let:

$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$

be homomorphisms.


Then the composite of $\phi$ and $\psi$ is also a homomorphism.


Proof

So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.

Then what we are trying to prove is denoted:

$\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is a homomorphism.


To prove the above is the case, we need to demonstrate that the morphism property is held by $\circ$ under $\psi \bullet \phi$.


We take two elements $x, y \in G_1$, and put them through the following wringer:

\(\ds \map {\paren {\psi \bullet \phi} } {x \circ y}\) \(=\) \(\ds \map \psi {\map \phi {x \circ y} }\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map \psi {\map \phi x * \map \phi y}\) Definition of Morphism Property: applied to $\circ$ under $\phi$
\(\ds \) \(=\) \(\ds \map \psi {\map \phi x} \oplus \map \psi {\map \phi y}\) Definition of Morphism Property: applied to $*$ under $\psi$
\(\ds \) \(=\) \(\ds \map {\paren {\psi \bullet \phi} } x \oplus \map {\paren {\psi \bullet \phi} } y\) Definition of Composition of Mappings


Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\circ$ under $\psi \bullet \phi$.


Thus $\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is indeed a homomorphism.

$\blacksquare$


Sources