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!

Answer 1

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 $\mathbb{Z}_4$.

Answer 2

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