Consider subcategory of $Ab$ of category abelian groups consisting divisible abelian groups (such groups $X$ that for any $x \in X$ and $n \in \mathbb N$ there is $y \in X$ such that $ x = ny$).
I need to show that morphism $\varphi: \mathbb R \to \mathbb T$, $t \mapsto e^{it} $ is monomorhism. It is probably easy so I'm asking for just hint.
P.s. I don't get why $\mathbb T$ is said to be divisible? It is so if we define action of $n$ as $n\cdot e^{it}: = e^{i(t+n)}$, though.
Concerning the divisibility of ${\mathbb T}$: The action of ${\mathbb Z}$ on ${\mathbb T}$ is uniquely determined by the abelian group structure on ${\mathbb T}$ (it is for any abelian group) by $n.\exp(it)=\exp(it)^n=\exp(itn)$. In particular, ${\mathbb T}$ being divisible means that arbitrary roots exist in ${\mathbb T}$. Can you check that for yourself? Alternatively, note that it suffices to check divisibility for ${\mathbb R}$, because images of divisible groups are again divisible, and $\exp: {\mathbb R}\to{\mathbb T}$ is surjective.
Concerning the actual question: Consider first a morphism $\psi: X\to {\mathbb R}$ of a divisible abelian group $X$ into ${\mathbb R}$ such that $\exp\circ\psi$ is the trivial homomorphism. Then, looking at it as a morphism in the category of all abelian groups, it factors through the kernel of $\exp$. How does that kernel look like? Is it divisible?