0\longrightarrowAl{\overset{q}{\longrightarrow}}Bl{\overset{r}{\longrightarrow}}C\longrightarrow0.
If any of these statements holds, the sequence is called a split exact sequence, and the sequence is said to split.
In the above short exact sequence, where the sequence splits, it allows one to refine the first isomorphism theorem, which states that:
(i.e., isomorphic to the coimage of or cokernel of)to:
where the first isomorphism theorem is then just the projection onto .
It is a categorical generalization of the rank–nullity theorem (in the form in linear algebra.
First, to show that 3. implies both 1. and 2., we assume 3. and take as the natural projection of the direct sum onto, and take as the natural injection of into the direct sum.
To prove that 1. implies 3., first note that any member of B is in the set . This follows since for all in, ; is in, and is in, since
Next, the intersection of and is 0, since if there exists in such that, and, then ; and therefore, .
This proves that is the direct sum of and . So, for all in, can be uniquely identified by some in, in, such that .
By exactness . The subsequence implies that is onto; therefore for any in there exists some such that . Therefore, for any c in C, exists k in ker t such that c = r(k), and r(ker t) = C.
If, then is in ; since the intersection of and, then . Therefore, the restriction is an isomorphism; and is isomorphic to .
Finally, is isomorphic to due to the exactness of ; so B is isomorphic to the direct sum of and, which proves (3).
To show that 2. implies 3., we follow a similar argument. Any member of is in the set ; since for all in,, which is in . The intersection of and is, since if and, then .
By exactness,, and since is an injection, is isomorphic to, so is isomorphic to . Since is a bijection, is an injection, and thus is isomorphic to . So is again the direct sum of and .
An alternative "abstract nonsense" proof of the splitting lemma may be formulated entirely in category theoretic terms.
In the form stated here, the splitting lemma does not hold in the full category of groups, which is not an abelian category.
It is partially true: if a short exact sequence of groups is left split or a direct sum (1. or 3.), then all of the conditions hold. For a direct sum this is clear, as one can inject from or project to the summands. For a left split sequence, the map gives an isomorphism, so is a direct sum (3.), and thus inverting the isomorphism and composing with the natural injection gives an injection splitting (2.).
However, if a short exact sequence of groups is right split (2.), then it need not be left split or a direct sum (neither 1. nor 3. follows): the problem is that the image of the right splitting need not be normal. What is true in this case is that is a semidirect product, though not in general a direct product.
To form a counterexample, take the smallest non-abelian group, the symmetric group on three letters. Let denote the alternating subgroup, and let . Let and denote the inclusion map and the sign map respectively, so that
0\longrightarrowAl{\stackrel{q}{\longrightarrow}}Bl{\stackrel{r}{\longrightarrow}}C\longrightarrow0
is a short exact sequence. 3. fails, because is not abelian, but 2. holds: we may define by mapping the generator to any two-cycle. Note for completeness that 1. fails: any map must map every two-cycle to the identity because the map has to be a group homomorphism, while the order of a two-cycle is 2 which can not be divided by the order of the elements in A other than the identity element, which is 3 as is the alternating subgroup of, or namely the cyclic group of order 3. But every permutation is a product of two-cycles, so is the trivial map, whence is the trivial map, not the identity.