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I am trying to find whether $\{1\}\longrightarrow\mathbb{Z}\longrightarrow\mathbb{R}\longrightarrow\mathbb{R}/\mathbb{Z}\longrightarrow \{1\}$ splits. My conjecture is it is not as we cannot find a non-zero group homomorphism from $\mathbb{R}/\mathbb{Z}$ to $\mathbb{R}$. If my conjecture is correct, can we tweak the above sequence so that it splits?

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    $\begingroup$ By the way with abelian groups, denoting the operation by $+$, it's typical to refer to the trivial group as $\{0\}$. $\endgroup$ – Sammy Black Oct 13 '14 at 2:55
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Is not split:

If it is we get $\Bbb R=\Bbb R/\Bbb Z\oplus \Bbb Z$ which is not possible, because there is no element of finite order in $\Bbb R$.

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  • $\begingroup$ @ Hamou $\Bbb R$ is connected, but $\Bbb Z$ is not connected. $\endgroup$ – aliakbar Aug 19 '15 at 4:51
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The sequence $0 \to \mathbb{Q}\to \mathbb{R} \to \mathbb{R}/\mathbb{Q}\to 0$ splits.

To have a split one you need the subgroup to be divisible ( a $\mathbb{Q}$ vector subspace).

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  • $\begingroup$ Oh, so $\Bbb R \cong \Bbb Q \oplus \Bbb R / \Bbb Q$? Interesting... $\endgroup$ – Matthew Levy Dec 23 '14 at 21:55
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    $\begingroup$ A subgroup of a $\mathbb{Q}$-vector space is a direct summand if and only if it is in fact a subspace, that is, a divisible subgroup. Note that $\mathbb{R}$ is a $\mathbb{Q}$ vector space. $\endgroup$ – Orest Bucicovschi Dec 24 '14 at 4:38

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