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Exercise 8.24 in Aluffi's Algebra: Chapter 0 asks us to find an epimorphism in $\text{Grp}$ without right inverses.

I happen to know that epimorphisms in $\text{Grp}$ are surjective, so we need a surjective $G\xrightarrow{\phi}G'$ without right inverses. But $G'\cong G/\operatorname{ker}(\phi)$, so in a sense we need a $G/\operatorname{ker}(\phi)$ that cannot be realised as a subgroup of $G$.

I tried something but failed. Can someone give a hint?

Thanks!

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    $\begingroup$ Try to play a bit with the quaternion group and check which possible subgroups and quotients it has. $\endgroup$ – Tobias Kildetoft Dec 11 '12 at 14:56
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The Quaternion Group $Q_8$ has a center of order $2$, and $Q_8/Z(Q_8)$ is isomorphic to the Klein $V$ group. However every subgroup of $Q_8$ of order $4$ is isomorphic to $\mathbb{Z}_4$.

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Consider the epimorphism $\phi:\mathbb Z\to \mathbb Z_{2},\phi(n)=n \pmod 2$.

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  • $\begingroup$ But this one has an right inverse. I guess we cannot find what we need in abelian groups. $\endgroup$ – Hui Yu Dec 11 '12 at 15:22
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    $\begingroup$ No, it does not have a right inverse (since the integers does not contain any element of order $2$, and any right inverse must be injective). $\endgroup$ – Tobias Kildetoft Dec 11 '12 at 15:29
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    $\begingroup$ In fact, surjections between cyclic groups provide lots of examples. $\endgroup$ – Mariano Suárez-Álvarez Mar 29 '13 at 11:36

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