An epimorphism in $\text{Grp}$ without right inverse?

Question

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!

Try to play a bit with the quaternion group and check which possible subgroups and quotients it has. — Tobias Kildetoft, Dec 11 '12 at 14:56

score 5 · Answer 1 · edited Mar 29 '13 at 11:23

5

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

edited Mar 29 '13 at 11:23

answered Dec 11 '12 at 15:21

Aliakbar

3,167
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But this one has an right inverse. I guess we cannot find what we need in abelian groups. – Hui Yu Dec 11 '12 at 15:22
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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). – Tobias Kildetoft Dec 11 '12 at 15:29
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In fact, surjections between cyclic groups provide lots of examples. – Mariano Suárez-Álvarez Mar 29 '13 at 11:36

score 3 · Accepted Answer · answered Dec 11 '12 at 14:59

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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 $\mathbb{Z}_4$.

answered Dec 11 '12 at 14:59

Alexander Gruber

26,963

An epimorphism in $\text{Grp}$ without right inverse?

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