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I was reading a proof to this proposition from this link:

http://planetmath.org/abeliangroupisdivisibleifandonlyifitisaninjectiveobject

they proceed by contradiction, so $A$ is not divisible and there exists $a \in A$ and $n \in \mathbb{N}$ such that the equation $nx=a$ does not have solution in $A$. Then they defined $B = <a>$ and they considered two cases, when $B$ is infinite and when $B$ is finite but I don't understand the case when the order of the group $B$ is finite. In particular I don't know how to prove that $n$ does not divide $k$ and how to use it to define the homomorphism $f$ such that $f(a)=n$.

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The assertion that $n$ does not divide $k$ is incorrect, and is not used anywhere in the rest of the argument.

The homomorphism $f$ is defined by $f(ma)=mn$ for each integer $m$. This is well-defined since $ma=m'a$ iff $m-m'$ is divisible by $k$ (since $a$ has order $k$), but then $mn=m'n$ in $\mathbb{Z}_{nk}$ since $mn-m'n=(m-m')n$ is divisible by $nk$.

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  • $\begingroup$ Is your definition of $f$ equivalent to defining $f(a) = n$, and extending it to a homomorphism? If so, why do we need to check that $f$ is well-defined? $\endgroup$ – Joshua Ruiter Feb 4 '17 at 22:32
  • $\begingroup$ Well, how do you know that it is possible to extend $f(a)=n$ to a homomorphism? There is something you must check there, no matter how you phrase it. $\endgroup$ – Eric Wofsey Feb 4 '17 at 22:35
  • $\begingroup$ I guess you're right. I was thinking that you can always extend a map defined on a generating set to a homomorphism. But that's incorrect. The result I was thinking of says that if there is an extension to a homomorphism, then it's unique. $\endgroup$ – Joshua Ruiter Feb 4 '17 at 22:38

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