Can someone please check over the correctness of my solution for the following question:
Let $G$ be an Abelian group and $\phi:G\rightarrow G$ a homomorphism such that $\phi(\phi(g))=g$ for all $g\in G$. (such a homomorphism is called a $\textbf{projection}$ homomorphism) Show that $G\cong \phi(G) \times \text{ker }\phi$.
Hint: consider for all $g\in G$, $g-\phi(g)$ [or in direct product notation: $g\phi^{-1}(g)$]
The question came from the same source as the following question in this post: Show $G=H\oplus K$. The relevant definitions and theorems assumed will be the same as that post therein.
As suggested in the hint let $y=g(\phi(g))^{-1}$, then $\phi(y)=\phi(g(\phi(g))^{-1})=\phi(g)\phi((\phi(g))^{-1})=\phi(g)(\phi(\phi(g)))^{-1}=\phi(g)g^{-1}$, because $\phi$ is a homomorphism and $(\phi(g))^{-1}=\phi(g^{-1})$.
Let $g\in \text{ker }\phi$ implies $\phi(g)=e$ and $\phi(\phi(g))=g=\phi(e)=e$, and so $g=e$, this means that $\text{ker }\phi=\{e\}$ and so $\phi$ is injective.
Since $\phi$ is one to one, $\phi(\phi(g)g^{-1})=e$ then $\phi(\phi(g)g^{-1})\in \text{ker }\phi$, and $G$ is abelian, $g\phi(g^{-1})=g(\phi(g))^{-1}=e.$ Which also gives $\phi(g)=g$ for all $g\in G$. Hence $\phi$ is the identity map on all of $G$.
Also $g\in \phi(G)$, then $\phi(g)=g$. Since $\phi(\phi(g))=\phi(g)=g=e$, then $g\in \phi(G) \cap \text{ker }\phi={e}.$ Lastly, $g=g(\phi(g))^{-1}g$, because $g=\phi(g)=\phi(g((\phi(g))^{-1})g=\phi(g)\phi((\phi(g))^{-1})\phi(g)=\phi(g)\phi(\phi(g^{-1}))\phi(g)=(gg^{-1})g.$ As $\phi(g)\phi((\phi(g))^{-1})\in \text{ker }\phi$ and $\phi(g)\in \phi(G)$, and $\text{ker }\phi\cdot \phi(G)=G.$ We can conclude that $\phi(G)\times\text{ker }\phi \cong G$
Thank you in advance.