- there exists a map
t:B→Asuch thattqis the identity onA,- there exists a map
u:C→Bsuch thatruis the identity onC,Bis isomorphic to the direct sum ofAandC(or a semidirect product ofAbyCin the case of non-abelian groups), withqbeing the natural injection ofAandrbeing the natural projection ontoC.

The short exact sequence is called *split* if any the above statements hold.

To prove that (1) implies (3), first note that any member of *B* is in the set (ker *t* + im *q*). This follows since for all *b* in *B*, *b* = (*b* - *qt*(*b*)) + *qt*(*b*); *qt*(*b*) is obviously in im *q*, and (*b* - *qt*(*b*)) is in ker *t*, since

*t*(*b*-*qt*(*b*)) =*t*(*b*) -*tqt*(*b*) =*t*(*b*) - (*tq*)*t*(*b*) =*t*(*b*) -*t*(*b*) = 0.

(By exactness, im *q* = ker *r*, so im *q* is a normal subgroup of *B* in the case of groups.)

This proves that *B* is the direct sum (alternatively, a semidirect product) of im *q* and ker *t*. So, for all *b* in *B*, *b* can be uniquely identified by some *a* in *A*, *k* in ker *t*, such that *b* = *q*(*a*) + *k*.

By exactness, ker *rq* = *A*, and so ker *r* = im *q*. The subsequence *B* → *C* → 0 implies that *f* is onto; therefore for any *c* in *C* there exists some *b* = *q*(*a*) + *k* such that *c* = *r*(*b*) = *r*(*q*(*a*) + *k*) = *r*(*k*). Therefore, for any *c* in *C*, exists *k* in ker *t* such that *c* = *r*(*k*), and *r*(ker *t*) = *C*.

If *r*(*k*) = 0, then *k* is in im *q*; since the intersection of im *q* and ker *t* = 0, then *k* = 0. Therefore the restriction of the morphism *f* : ker *t* → *C* is an isomorphism; and ker *t* is isomorphic to *C*.

Finally, im *q* is isomorphic to *A* due to the exactness of 0 → *A* → *B*; so *B* is isomorphic to the direct sum (alt., a semidirect product) of *A* and *C*, which proves (3).

To show that (2) implies (3), we follow a similar argument. Any member of *B* is in the set ker *r* + im *u*; since for all *b* in *B*, *b* = (*b* - *ur*(*b*)) + *ur*(*b*), which is in ker *r* + im *u*. The intersection of ker *r* and im *u* is 0, since if *r*(*b*) = 0 and *u*(*c*) = *b*, then 0 = *ru*(*c*) = *c*.

By exactness, im *q* = ker *r*, and since *q* is an injection, im *q* is isomorphic to *A*, so *A* is isomorphic to ker *r*. Since *ru* is a bijection, *u* is an injection, and thus im *u* is isomorphic to *C*. So *B* is again the direct sum of *A* and *C*.
In general, however, not all A will give rise to a semidirect product, though, in the case of groups. See http://planetmath.org/encyclopedia/SemidirectProduct.html on this.