# Inverse image of surjective group homomorphism and section

Let $p : G \twoheadrightarrow H$ be a surjective homomorphism of groups.

Question 1: let $a,b \in H$, is $p^{-1}(ab) = p^{-1}(a)p^{-1}(b)$?

We do have that $p(p^{-1}(ab))=ab$ and $p(p^{-1}(a)p^{-1}(b)) = pp^{-1}(a)pp^{-1}(b) = a b$ but $p$ need not be injective. So I think the answer is "no" in general.

Question 2: I want to construct a section $f : H \to G$ such that $p \circ f = \mathrm{id}_H$, $f(ab)=f(a)f(b)$ and $f(1)=1$, that is: the section $f$ is an injective homomorphism. Does this section exist? And if it does, is it unique?

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How do you define $p^{-1}(a)p^{-1}(b)$? Is it $\{xy : p(x) = a,\, p(y) = b\}$? – Daniel Fischer Sep 3 '13 at 12:45
Yes. I also want to consider the case when $p^{-1}(a)$ means taking any preimage element, i.e. choose $x \in G$ such that $p(x)=a$ and define $"p^{-1}"(a)=x$. I formalized this notion in the second question which is more important: the existence and uniqueness of a section. – LinAlgMan Sep 3 '13 at 12:48
The answer for the first is affirmative, let $N = \ker p$. Then $p^{-1}(a)p^{-1}(b) = xNyN = xyNN = xyN = p^{-1}(ab)$ for all $x,\,y$ with $p(x) = a,\, p(y) = b$. – Daniel Fischer Sep 3 '13 at 13:00

## 3 Answers

Question 2 has a negative answer I'm afraid. The canonical projection

$$\mathbb Z\to\mathbb Z/2$$

is surjective, but the only group homomorphism

$$\mathbb Z/2\to\mathbb Z$$

is trivial.

Addendum: Yes, your question is strongly related to short exact sequences. If $K$ is the kernel of your surjective map $p$, then $$0\to K\to G\to H\to 0$$ is exact. For abelian groups we have now: A section like the one in your question (also called a splitting) exists, if and only if $G\cong K\oplus H$, if and only if there exists a splitting $G\to K$.

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Thank you. Can you say in what case, or in which additional conditions, it does exist? (I know the answer is related to en.wikipedia.org/wiki/Split_exact_sequence in some way). – LinAlgMan Sep 3 '13 at 13:00
You could also extend your comment to arbitrary groups, since then the section will still give you a splitting (ie, a semidirect product). – Tobias Kildetoft Sep 3 '13 at 13:22
@TobiasKildetoft, yes but I personally couldn't do much more than copy verbatim from wikipedia (which gives a satisfying answer). The only idea, which was sorta missing, is that any surjection gives rise to a short exact sequence. – Simon Markett Sep 3 '13 at 13:28

There are various conditions under which the short exact sequence $0 \to K \to G \to H \to 1$ splits, but nothing very general. For example, if $G$ is finite and $|K|$ and $|H|$ are coprime, then the extension splits by the Schur-Zassenhaus Theorem (which says also that all complements of $K$ in $G$ are conjugate in $G$).

Another sufficient condition is when $K$ is a complete group, which means that its centre is trivial and its outer automorphism group is trivial. So if, for example $K = S_n$ with $n \ge 3$ and $n \ne 6$, then the extension definitely splits. A complement in that case is $C_G(K)$.

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To address the uniqueness part of your Question 2, no, even when a section exists, it is not unique.

A couple of examples $G$.

The first is an abelian group, the Klein group $$G = \langle a, b : a^2 = b^2 = 1, ab = ba \rangle.$$ Take $H = \langle c \rangle$ cyclic of order $2$, and $p(a) = 1, p(b) = c$. Then there are two sections $f$, mapping $c$ respectively to $b$ and $a b$.

The second is the nonabelian group $S_{3}$, $$G = \langle a, b : a^3 = b^2 = 1, b^{-1}ab = a^{-1} \rangle.$$ $H$ is the same as above, $p(a) = 1, p(b) = c$. Here there are three sections $f$, mapping $c$ respectively to $b, ba, b a^{2}$.

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Thank you for the counter-examples. – LinAlgMan Sep 8 '13 at 8:23