Does this ses $0\rightarrow \mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{R}/p\mathbb{Z}\rightarrow \mathbb{R}/\mathbb{Z}\rightarrow 0$ split?

Question

Does the following short exact sequence of $\mathbb{Z}$-modules split ?

$$0\rightarrow \mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{R}/p\mathbb{Z}\rightarrow \mathbb{R}/\mathbb{Z}\rightarrow 0,$$

where all maps are the natural maps.

Answer 1

No, this sequence does not split. Indeed, a splitting would give rise to a morphism $\mathbb{R}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ which when post-composed with the map $\mathbb{Z}/p\mathbb{Z} \to \mathbb{R}/p\mathbb{Z}$ gives the identity on $\mathbb{Z}/p\mathbb{Z}$. But $\mathbb{R}$ is a divisible abelian group, and therefore so is any quotient, e.g. $\mathbb{R}/p\mathbb{Z}$. Hence, any group homomorphism $\mathbb{R}/p\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$ is the zero map. (Precompose with multiplication by $p$ on $\mathbb{R}/p\mathbb{Z}$, which is surjective.)

Answer 2

If the s.e.s. split, there would be a homomorphism: $$\theta\colon \mathbb{R}/\mathbb{Z}\to \mathbb{R}/p\mathbb{Z},$$ taking $\frac1p\in \mathbb{R}/\mathbb{Z}$ to a preimage $\alpha=\theta\left(\frac1p\right)\in\mathbb{R}/p\mathbb{Z}$. Then $\alpha$ would have the form $\frac1p+k$ for some integer $k$.

In particular $p\alpha=1+kp\neq 0$ in $\mathbb{R}/p\mathbb{Z}$. However: $$p\alpha=p\theta\left(\frac1p\right)=\theta(1)=\theta(0)=0,$$ yielding a contradiction. Therefore the s.e.s. does not split.