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Consider $\mathbb{R}$ and $\mathbb{Z}$ as groups with operation as an usual addition. Then does there exist an onto group homomorphism from $\mathbb{R}\to \mathbb{Z}$ ?

My intuition tells me that such homomorphism doesn't exist. So far , I have tried the following.

Suppose $f$ is such homomorphism. Then ,by first isomorphism theorem , $\mathbb{R}/Ker f \cong \mathbb{Z}$. I want to show that the above supposition is wrong. so I have to prove either $\mathbb{R}/Ker f$ is uncountable or it is noncyclic. But I don't know how to go further to prove any cases.

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Suppose $\phi: \mathbb{R} \to \mathbb{Z}$ is a group homomorphism, and let $x \in \mathbb{R}$, and let $n = \phi(x)$. Then \begin{align*} n &= \phi(x) \\ &= \phi\left((n+1)\left(\frac{x}{n+1}\right)\right) \\ &= (n+1) \phi\left( \frac{x}{n+1} \right). \end{align*} Therefore, $\phi\left( \frac{x}{n+1} \right) = \frac{n}{n+1}$. But this has to be an integer, so we conclude that $\boldsymbol{n=0}$. (The argument was technically incorrect for $n=-1$. We could use $n^2 + 1$ instead of $n+1$.)

In other words, $\phi(x) = 0$ for all $x \in \mathbb{R}$, since $x$ was arbitrary.

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Since $\mathbb{R}$ is divisible, any map $\mathbb{R} \to \mathbb{Z}$ must vanish.

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