gr.group theory - When does a subgroup H of a group G have a complement in G?

A lot can be said in the finitely generated abelian case, just by using the structure theorem.

Call a group transversal if every subgroup has a complement; non-transversal of it has proper, nontrivial subgroups, none of which has a complement; and semi-transversal if it is not transversal but some proper, nontrivial subgroup has a complement.

Take a finite abelian group $G=mathbb{Z}/p_1^{e_1}mathbb{Z}timesldotstimesmathbb{Z}/p_k^{e_k}mathbb{Z}$ with $rgeq 0$, distinct primes $p_1,ldots,p_k$ and $e_igeq 1$.

Then $G$ is transversal when $e_i=1$ for all $i$, semi-transversal when $k>1$ and $e_i>1$ for some $i$, and non-transversal when $k=1$ and $e_1>1$.

For an infinite finitely generated abelian group $G$, $G$ is non-transversal if $G=mathbb{Z}$, and semi-transversal otherwise.

Given a finitely generated abelian group $G$ and $Ssubseteq G$, call $S$ independent if $0notin S$ and for all $x_1,ldots,x_kin S$, $r_1,ldots, r_kinmathbb{Z}$, we have that $sum_{i=1}^k r_ix_i=0$ implies $r_ix_i=0$ for all $i$. Call $S$ a basis of $G$ if it is independent and $G=left< Sright>$.

Then if $Hleq G$, $H$ has a complement if and only if $H$ has a basis that can be expanded to a basis of $G$. If $S$ is a basis for $H$ and $Ssubseteq T$ for some basis $T$ of $G$, then the complement of $H$ is $left< Tbackslash Sright>$.

So if $G$ is a vector space over $mathbb{Z}/pmathbb{Z}$ then any subgroup has a complement, since any subspace has a basis that can be completed to the whole space. My definition of independence is the same as linear independence.

If $G$ is a free module over $mathbb{Z}/p^nmathbb{Z}$ then my definition of independence is weaker than linear independence. I would like to say that $Hleq G$ has a complement if and only if it has a module basis, but I can't prove the reverse direction of this.

I know less about the divisible abelian case, except that if $G$ is a divisible abelian group then saying that $Hleq G$ has a complement is the same as saying that $H$ is divisible. In particular $mathbb{Q}$ and the Prufer $p$-group $mathbb{Z}(p^infty)$ are both non-transversal.