Finite groups $A$ and $B$. Does existence of surjective group homomorphism $f:A\to B$ imply existence of an injective homomorphism $g:B\to A$? 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
 A: 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.
A: Not in general. Like you said, the first isomorphism theorem says that $A/\ker f\cong B$. In particular, if $B$ injects into $A$, then so does the quotient $A/\ker f$.
Let's see if we can break this. Since normal subgroups are in bijection with kernels of maps out of $A$, it suffices to find a group $A$ and a normal subgroup $N \trianglelefteq A$ such that $A$ doesn't contain a copy of the quotient $A/N$.
Consider the quaternions $Q_8$. Its center, $Z = \{\pm 1\}$ is normal, but when you quotient by $Z$, you get $Q_8/Z \cong V$, the Klein 4-group. All order four subgroups of $Q_8$ are cyclic (generated by $i$, $j$, or $k$), so $Q_8$ doesn't contain this quotient.
