Let $G$ be a divisible group, written additively. Show that for every positive integer $n$, the function $\alpha : G → G$ given by $\alpha(a) = na$ is an onto homomorphism. Is it necessarily an automorphism?
The proof is very easy. We have $\alpha(a +b) = n(a +b) = na +nb = \alpha(a)+\alpha(b)$, so $\alpha$ is a homomorphism. If $a \in G$, then since $G$ is divisible, there exists a $b \in G$ such that $nb = a$. Thus, $\alpha(b) = a$, and $\alpha$ is onto.
But it is not necessarily an isomorphism. I found this counterexample, but here comes my problem, I could not understand it.
Let $G$ be the Prüfer $p$-group and take $n = p$. Then we see that $1/p + \mathbb{Z} \in \text{ker} (\alpha)$.
Why exactly $1/p + \mathbb{Z} \in \text{ker} (\alpha)$?
Btw, Prüfer $p$-group$:=\{a/p^n+\mathbb{Z} : a, n\in \mathbb{Z}, n\geq 0\}$.
$\alpha(1/p + \mathbb Z) = p(1/p + \mathbb Z)$. Recall that the sum of cosets $(a + \mathbb Z) + (b + \mathbb Z)$ is defined as $(a+b) + \mathbb Z$. Thus, $p(1/p + \mathbb Z) = 1 + \mathbb Z = 0 + \mathbb Z$ which is the identity element of the Prüfer $p$-group. Hence, $1/p + \mathbb Z \in ker(\alpha)$.