For finite groups $A$ and $B$. Does existence of a surjective group homomorphism $f:A\to B$ imply the existence of an injective group homomorphism $g:B\to A$?
I know that since $f$ is surjective we have $\frac{A}{\ker(f)}\cong B$
If I take an element $b\in B$, I know $b=f(a)$ for some $a\in A$.
I want to send $f(a)\mapsto a$, but of course there are multiple options since the pre-image of $f(a)$ may be larger than just the set $\{a\}$.
Is there a more natural way to find an injection? or is the claim false?
Thanks
Best Answer
No, it does not. Let $A=Q_8$ denote the quaternion group, and let $B=C_2\times C_2$. Since $Q_8$ modulo its center $Z(Q_8)=\{\pm 1\}$ is isomorphic to $B$, it follows that there is a surjective group homomorphism from $A$ to $B$ with kernel $\{\pm 1\}$. On the other hand, there is no injective group homomorphism from $B$ to $A$ because every subgroup of $Q_8$ of order $4$ is cyclic, unlike $B$.
The same example appears here and likely appears already in some form on this website.