Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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 3

up vote 3 down vote accepted

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 $G\to K$.

share|improve this answer
    
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 $C_G(K)$.

share|improve this answer

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 $b, ba, b a^{2}$.

share|improve this answer
    
Thank you for the counter-examples. –  LinAlgMan Sep 8 '13 at 8:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.