# 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

• The title doesn't say it all. Morphisms of what? Groups? Also, put the question in the body of the question, not just the title. – Captain Lama Dec 11 '19 at 5:04
• I have made some edits – Jungleshrimp Dec 11 '19 at 5:06
• Can you find an injective group homomorphism between the additive groups $\mathbb Z/2\mathbb Z\to\mathbb Z$? – trisct Dec 11 '19 at 5:08
• I forgot to include the assumption that both groups are finite... – Jungleshrimp Dec 11 '19 at 5:15
• I have taken the liberty of fixing the notation by changing $f:A\mapsto B$ to $f:A\to B.$ The arrows $\text{“} to \text{''}$ and $\text{“} \mapsto \text{''}$ have different meanings. The latter is used in things like $x\mapsto (x+2y)^3,$ which is a different function from $y\mapsto (x+2y)^3. \qquad$ – Michael Hardy Dec 11 '19 at 5:41

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 $$B$$.
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.