Prove Function Composition's Isomorphism

Question

Dr. Pinter's "A Book of Abstract Algebra" presents this exercise:

Given the isomorphic functions: $$f: G_{1}\rightarrow G_{2}$$ $$g: G_{2}\rightarrow G_{3}$$

Prove that $g\cdot f$ is isomorphic for $G_{1}\rightarrow G_{3}$.

Claims: $g\cdot f$ is bijective, i.e. injective and surjective.

Proof: From Chapter 6's (Functions):

If $f$ and $g$ are bijective, then $g\cdot f$ is bijective.

Claim: $(g\cdot f)(ab) = (g\cdot f)(a)\cdot(g\cdot f)(b)$

Proof: By Chapter 6's (Functions) definition of a composite function:

$[g\cdot f](x) = g(f(x))$

Can $ab$ be substituted for $x$ to complete this proof?

Answer 1

I don't really get your question, but for the second claim write it down:

$$(g \cdot f )(a*b) = g ( f(a*b) ) = g(f(a)*f(b)) = g(f(a))*g(f(b)) = (g\cdot f)(a)* (g\cdot f)(b)$$

where I indicate $*_1, *_2, *_3$ with $*$ because it's not important at all

Answer 2

The $x$ is just a placeholder for the parameter of the composition function which you denote by $(g\cdot f)$ and in the second occurrence as $[g\cdot f]$. So applying this function on an 'actual value' like $ab$ is done by replacing the placeholder with the value.

Just like, define $f(x):= x^4$ then by definition $f(2) = 2^4 = 16$ or $f(ab)= a^4b^4$. So yes you can substitute.

It does not complete the proof. You did not prove that the composition $(g\cdot f)$ is a homomorphism, which means that $(g\cdot f)(ab)=(g\cdot f)(a)(g\cdot f)(b)$ holds for arbitrary $a,b$.