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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

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  • $\begingroup$ 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. $\endgroup$ Dec 11, 2019 at 5:04
  • $\begingroup$ I have made some edits $\endgroup$ Dec 11, 2019 at 5:06
  • $\begingroup$ Can you find an injective group homomorphism between the additive groups $\mathbb Z/2\mathbb Z\to\mathbb Z$? $\endgroup$
    – trisct
    Dec 11, 2019 at 5:08
  • $\begingroup$ I forgot to include the assumption that both groups are finite... $\endgroup$ Dec 11, 2019 at 5:15
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    $\begingroup$ 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 $y\mapsto (x+2y)^3. \qquad$ $\endgroup$ Dec 11, 2019 at 5:41

2 Answers 2

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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.

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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.

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