To show that if $f: G \to H$ and $g: H \to K$ are morphisms, then $g \circ f: G \to K$ is also a morphism, we'll proceed as follows.

Definitions and Assumptions

• Let $G$, $H$, and $K$ be groups.
• A morphism (or homomorphism) between two groups is a function that preserves the group operation. Specifically, if $f: G \to H$ is a morphism, then for all $x, y \in G$, we have: $f(xy) = f(x)f(y)$ Similarly, if $g: H \to K$ is a morphism, then for all $a, b \in H$, we have: $g(ab) = g(a)g(b)$

Proof

We need to show that the composition $g \circ f: G \to K$ is a morphism, i.e., for all $x, y \in G$: $(g \circ f)(xy) = (g \circ f)(x)(g \circ f)(y)$

1. Consider the composition $g \circ f: G \to K$ defined by: $(g \circ f)(x) = g(f(x)) \quad \text{for all } x \in G$

2. Evaluate the composition on the product $xy$: $(g \circ f)(xy) = g(f(xy))$

3. Use the fact that $f$ is a morphism: Since $f$ is a morphism, we have: $f(xy) = f(x)f(y)$ Therefore: $(g \circ f)(xy) = g(f(x)f(y))$

4. Use the fact that $g$ is a morphism: Since $g$ is a morphism, we have: $g(f(x)f(y)) = g(f(x))g(f(y))$ Therefore: $(g \circ f)(xy) = g(f(x))g(f(y))$

5. Recognize the result: Notice that: $g(f(x)) = (g \circ f)(x) \quad \text{and} \quad g(f(y)) = (g \circ f)(y)$ Hence: $(g \circ f)(xy) = (g \circ f)(x)(g \circ f)(y)$

Conclusion

Since $(g \circ f)(xy) = (g \circ f)(x)(g \circ f)(y)$ for all $x, y \in G$, the composition $g \circ f$ is a morphism from $G$ to $K$.

Would you like further details or have any questions?

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Here are 5 questions that relate to this topic:

1. What are some examples of group homomorphisms?
2. How can the concept of morphism be extended to other algebraic structures like rings?
3. What is the kernel of a group homomorphism, and how does it relate to the morphism's injectivity?
4. Can you explain the concept of an isomorphism and how it relates to homomorphisms?
5. How does the First Isomorphism Theorem relate to group homomorphisms?

Tip: When composing two group homomorphisms, always remember that the composition of homomorphisms is itself a homomorphism. This property can greatly simplify working with complex algebraic structures.