Splitting Lemma - Non-abelian Groups - Counterexample


To form a counterexample, take the smallest non-abelian group, the symmetric group on three letters. Let A denote the alternating subgroup, and let . Let q and r denote the inclusion map and the sign map respectively, so that

is a short exact sequence. Condition (3) fails, because is not abelian. But condition (2) holds: we may define u: CB by mapping the generator to any two-cycle. Note for completeness that condition (1) fails: any map t: BA 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 A is the alternating subgroup of, or namely the cyclic group of order 3. But every permutation is a product of two-cycles, so t is the trivial map, whence tq: AA is the trivial map, not the identity.


Mac Lane, S. Homology. Reprint of the 1975 edition. Springer Classics in Mathematics ISBN 3-540-58662-8

Read more about this topic:  Splitting Lemma, Non-abelian Groups

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