Editing Splitting lemma (section)

===Counterexample===
To form a counterexample, take the smallest [[non-abelian group]] {{math|''B'' ≅ ''S''{{sub|3}}}}, the [[symmetric group]] on three letters.  Let {{math|''A''}} denote the [[alternating group|alternating subgroup]], and let {{math|''C'' {{=}} ''B''/''A'' ≅ {±1}}}.  Let {{math|''q''}} and {{math|''r''}} denote the inclusion map and the [[parity of a permutation|sign]] map respectively, so that

: <math>0 \longrightarrow A \mathrel{\stackrel{q}{\longrightarrow}} B \mathrel{\stackrel{r}{\longrightarrow}} C \longrightarrow 0 </math>

is a short exact sequence.  3. fails, because {{math|''S''{{sub|3}}}} is not abelian, but 2. holds: we may define {{math|''u'': ''C'' → ''B''}} by mapping the generator to any [[cyclic permutation|two-cycle]].  Note for completeness that 1. fails: any map {{math|''t'': ''B'' → ''A''}} must map every two-cycle to the [[identity permutation|identity]] because the map has to be a [[group homomorphism]], while the [[order (group theory)|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 {{math|''A''}} is the alternating subgroup of {{math|''S''{{sub|3}}}}, or namely the [[cyclic group]] of [[order of a group|order]] 3. But every [[permutation]] is a product of two-cycles, so {{math|''t''}} is the trivial map, whence {{math|''tq'': ''A'' → ''A''}} is the trivial map, not the identity.