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Lemma·L39

Kernels are normal subgroups

The kernel of any homomorphism is a normal subgroup.

For every homomorphism , the kernel is a normal subgroup of :
In words
For every homomorphism phi from a group G to a group H, its kernel is a normal subgroup of the source group G: not merely a subgroup, it is unchanged under conjugation by any element of the group.
Never needed: F03 · F04 · F06 · F08 · F10 · F11 · F12 · A03 · A04 · A05 · A06 · A07 · A08 (computed from the citation graph, not asserted).

Proof

  1. 1
    is a subgroup of by L38 (i); it remains to verify conjugation-invariance (D048).
  2. 2
    Let and , so (D047). Then using the homomorphism property (applied twice, with associativity), , L37 (ii) (so ), and D038 (G2, G3) in . Hence (D047), and therefore .

Remarks

This is the easy half of the correspondence between normal subgroups and homomorphisms. The converse, that every normal subgroup is the kernel of some homomorphism (its quotient map), needs the quotient-group construction; together they yield the first isomorphism theorem .