Splitting lemma

and 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. ==Non-abelian groups==

In the form stated here, the splitting lemma does not hold in the full category of groups, which is not an abelian category. Partially true 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. Counterexample To form a counterexample, take the smallest non-abelian group , the symmetric group on three letters. Let denote the alternating subgroup, and let {{math|C B/A ≅ {±1}}}. Let and denote the inclusion map and the sign map respectively, so that : 0 \longrightarrow A \mathrel{\stackrel{q}{\longrightarrow}} B \mathrel{\stackrel{r}{\longrightarrow}} C \longrightarrow 0 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. ==References==