Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A remark from Wikipedia,

(about an exact sequence) $0\to A\to B\to C\to 1$

says " if a short exact sequence of groups is right split ... then it need not be left split or a direct sum ... what is true in this case is that B is a semidirect product, though not in general a direct product."

I cannot verify this, in particular showing that $AC = B$ is giving me problems as I don't know how to work with the right inverse (injection?) map $u:C\to B$.

If I define the map $\psi:AC\to B$ by $(a,c)\mapsto au(c)$, I want to show it is a bijection. But I can't even show it is well-defined. Is this the right bijection?

share|cite|improve this question
Actually I think I figured this out. The AC = B axiom anyway. – roo Oct 14 '12 at 23:18
I was wrong, I was able to show the map is well-defined though using the fact that $f:B\to C$ satisties $f\circ u = id_{C}$. I wasn't able to show that the map is injective or surjective though. – roo Oct 14 '12 at 23:29
OK! I got surjectivity. I had to use the isomorphism between $C$ and $B/A$. – roo Oct 15 '12 at 0:17
Dear lovinglife, Given that you have the map $u$, the map $(a,c) \mapsto a u(c)$ is certainly well-defined (although it may not be immediately obvious that it has all the other properties you need). What makes you concerned that it is not well-defined? Regards, – Matt E Oct 15 '12 at 0:39
Sorry I was confusing maps. For the map I said above, injectivity is what I was really seeking. Actually this confusion was what was holding me back. Surjectivity was the valuable exercise though. – roo Oct 15 '12 at 17:36
up vote 2 down vote accepted

Call the maps $f:A \rightarrow B$ and $g:B \rightarrow C$. We have $B/f(A) \cong C$ with isomorphism induced by $g$ whose inverse is given by composition of $u$ with natural projection. Let $b \in B$. Then $u(g(b))*f(A) = b*f(A)$ so $u(g(b))=b*f(a)$ for some $a \in A$. Then $f(a^{-1})*u(g(b)) = b$. This proves $B = f(A)*u(C)$. That $f(A)$ is normal in $B$ and that $f(A) \cap u(C) = 1$ follows easily from being a short exact sequence.

share|cite|improve this answer

Your Answer


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.