Show that for abelian groups G, H such that there exist homomorphisms $f:G \to H$ and $g:H \to G$ and $g \circ f$ is the identity – H can be expressed as a direct product of G and another abelian group K.
My first thought was to see the image of $f$ and that there is an isomorphism from $g$ to $f(G)$.
Given that this is an abelian group, the subgroup is normal and so I thought that perhaps $G\ X\ H/G$ would be isomorphic to $H$ but according to the following page this approach is not entirely correct.
https://groupprops.subwiki.org/wiki/Question:Normal_subgroup_quotient_group_determine_whole_group
I am not sure what to do – should I construct a function explicitly $s.t.$ it takes the image of $f$ in $H$ through g and everything else not contained in the image to (1, h)? Also what if $H$ and $G$ are not commutative?
Best Answer
Let's use different notation: $\phi:A\to B$ and $\psi:B\to A$ homomorphisms of Abelian groups with $\psi\circ\phi=\text{id}_A$. I claim that $B$ is the direct sum of $\phi(A)$ and $\ker\psi$.
If $b\in B$, then $\psi(b-\phi(\psi(b)))=\psi(b)-\psi(\phi(\psi(b)))= \psi(b)-\psi(b)=0$ (using $\psi\circ\phi=\text{id}_A$). Therefore $b-\phi(\psi(b))\in\ker\psi$, and $b$ is the sum of an element of $\phi(A)$ (namely $\phi(\psi(b))$) and an element of $\ker\psi$ (namely $b-\phi(\psi(b))$).
If $b\in \phi(A)\cap\ker \psi$, then $\psi(b)=0$ and $b=\phi(a)$ for some $a$. Therefore $0=\psi(\phi(a))=a$ and so $b=\phi(a)=0$.
As $\phi(A)+\ker\psi=B$ and $\phi(A)\cap\ker\psi=\{0\}$ then $B=\phi(A)\oplus\ker\psi$.