# Right Split Exact Sequence implies Semi-Direct Product - Why?

A remark from Wikipedia, http://en.wikipedia.org/wiki/Splitting_lemma

(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?

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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

## 1 Answer

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.

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