Composition of Homomorphisms/Algebraic Structure

Theorem

Let:

• $\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$
• $\left({S_2, *_1, *_2, \ldots, *_n}\right)$
• $\left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$

be algebraic structures.

Let:

• $\phi: \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_2, *_1, *_2, \ldots, *_n}\right)$
• $\psi: \left({S_2, *_1, *_2, \ldots, *_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$

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:

$\left({\psi \bullet \phi}\right): \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$ is a homomorphism.

To prove the above is the case, we need to demonstrate that the morphism property is held by each of the operations $\circ_1, \circ_2, \ldots, \circ_n$ under $\psi \bullet \phi$.

Let $\circ_k$ be one of these operations.

• We take two elements $x, y \in S_1$, and put them through the following wringer:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\psi \bullet \phi}\right) \left({x \circ_k y}\right)\)	\(=\)	\(\displaystyle \psi \left({\phi \left({x \circ_k y}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of composition of mappings
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \psi \left({\phi \left({x}\right) *_k \phi \left({y}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Morphism Property applied to $\circ_k$ under $\phi$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \psi \left({\phi \left({x}\right)}\right) \oplus_k \psi \left({\phi \left({y}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Morphism Property applied to $*_k$ under $\psi$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\psi \bullet \phi}\right) \left({x}\right) \oplus_k \left({\psi \bullet \phi}\right) \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	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 $\circ_k$ under $\psi \bullet \phi$.

As this holds for any arbitrary operation $\circ_k$ in $\left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right)$, it follows that it holds for all of them.

Thus $\left({\psi \bullet \phi}\right): \left({S_1, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({S_3, \oplus_1, \oplus_2, \ldots, \oplus_n}\right)$ is indeed a homomorphism.

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 7.2$: Lemma $\text{(iv)}$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 12$: Theorem $12.4$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 60$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $8.3 \ \text{(i)}$