The question is as follows:
Let $G$ be an Abelian group and suppose that $A$ is a divisible subgroup of finite index. Show that $G = A \dot{\times} B$ for some $B \leq G$.
$\textbf{Some definitions and notations:}$
An Abelian group $A$ is said to be divisible if, for every $a \in A$ and positive integer $n$, there exists $b \in A$ such that $b^n = a$.
If $G$ is any group having two normal subgroups $M$ and $N$ with $MN = G$ and $M \cap N = 1$, we say that $G$ is the internal direct product of its subgroups $M$ and $N$. And we write $G = M \dot{\times} N$.
$\textbf{An idea for to prove it}:$
It is possible to prove that a group $D$ is divisible if and only if it satisfies the following “injectivity” condition: Any homomorphism $f : A \to D$ from any group $A$ into $D$ extends to any group $B$ which contains $A$, i.e., there exists a homomorphism $\bar{f} : B → D$ so that $f\bar{f}|_A = f$.
And then if we can prove this, then for $D \leq G$ divisible, then $D$ will be split, i.e $D$ will have a complement $H$ so that $G = D \dot{\times} H$. Because since $D$ is divisible, it is injective. Thus the identity mapping $D \to D$ extends to a homomorphism $r : G \to D$. Then $H = \ker r$ is a complement for $D$ in $G$.
Can you please let me know if I am wrong?
And if it is nice approach, can you please help me to complete it? (Please feel free to post a proof.)
Thanks!
$\textbf{Proposed proof:}$
Here I will post my proof for this question. Please let me know if I am wrong and please post the edited version as an answer or please post your own proof.
I appreciate your help a lot!
Thanks!
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$\textbf{Proof:}$
First we start by claiming that a group $A$ is divisible if and only if it satisfies the following “injectivity” condition: Any homomorphism $f : L \to A$ from any group $L$ into $A$ extends to any group $H$ which contains $L$, i.e., there exists a homomorphism $\bar{f} : H → A$ so that $f\bar{f}|_L = f$.
For to prove this claim, We use Zorn’s Lemma. Consider the partially ordered set $P$ of all pairs $(C, g)$ where $C$ is a subgroup of $H$ containing $L$ and $g : C \to G$ is an extension of $f$. Let $(C, g) \leq (A, h)$ if $C \leq A$ and $g = h\mid_C$. The set $P$ is nonempty since it contains $(L, f)$. Also any tower $(C_{\alpha}, g_{\alpha})$ in $P$ has an upper bound $(\cup C_\alpha,\cup g_{\alpha}) \in P$. Consequently, by Zorn’s Lemma, $P$ has a maximal element, say $(C, g)$.
We claim that $C = H$ and $g$ is the desired extension of $f$ to $H$.
To see this suppose $C < H$. Then there exists an $x \in H$ so that $x \notin C$. There are two cases Either $x + C$ has finite order in $H/C$ or it has infinite order.
In the second case, $<C,x> = C \dot{\times} <x>$ so $g : C \to G$ can be extended to $g \dot{\times} 0 : C \dot{\times} <x> \to G$ contradicting the maximality of $(C, g)$.
In the first case, let $n$ be the order of $x+C$ in $H/C$, i.e., $n > 0$ is smallest positive integer so that $nx \in C$. Since $G$ is divisible there is a $z \in G$ so that $nz = g(nx)$. By the Pasting lemma, $g : C \to G$ can be extended to the homomorphism $g \dot{\times} h : C \dot{\times} <x> \to G$ where $h : <x> \to G$ is given by $h(x) = z$. This contradicts the maximality of $(C, g)$. Therefore, divisible groups are injective.
Conversely, if $G$ is injective then it is obviously divisible: given any $x \in G$ and $n > 0$ let $f : n\mathbb{Z} \to G$ be given by $f(n) = x$ . If $G$ is injective this extends to a homomorphism $\bar{f} : \mathbb{Z} \to G$. But then $n\bar{f}(1) = f(n) = x$.
Now we claim that any divisible subgroup of $G$ splits, i.e., if $A \leq G$ is divisible then $A$ has a complement $B$ so that $G = A \dot{\times} B$.
For to prove this we see that since $A$ is divisible, it is injective. Thus the identity mapping $A \to A$ extends to a homomorphism $r : G \to D$. Then $B = \ker r$ is a complement for $A$ in $G$.
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Please let me know if my proof is not correct?
And please post what you think which is correct.
Thanks!