To show a composition of two maps, $f, g$ is an automorphism I need to show they are bijective and homomorphism. The homomorphism part is easy: If we have $f, g: G \rightarrow G$, then for some element $a,b \in G$ we get
$$ f \circ g (ab) = f(g(ab)) = f(g(a)g(b))$$
as $g4$ is an isomorphism. Then as $f$ is an isomorphism, we also get
$$f(g(ab)) = f \circ g(a) f \circ g(b)$$
I'm pretty sure that's correct.
Now to prove for bijection, I first thought about checking to see if the kernel was trivial as that would show injection, $g(a,b) = (0,0)$ iff $a = b = 0$. So putting this in the composition gives us $f \circ g(0,0) = f(g(0,0)) = f(0,0) = (0,0)$, hence the map is injective. We see from the definition of the map that every element in it maps to some element in $G$ and so the map is also surjective and therefore bijective.
Is that correct?
Best Answer
The injectivity is easy in this case, $f(g(x))=0$ then $g(x)=0$ then $x=0$.
Surjectivity follows from the fact that for every $y$ there is some $x$ such that $f(x)=y$, and we can apply this to $x$ to obtain $u$ such that $g(u)=x$. In this case $f(g(u))=f(x)=y$.
Generally speaking, the composition of injections is an injection; and composition of surjections is a surjection; and therefore the composition of bijections is a bijection. Without requiring the additional structure.