Retract not implies normal complements are isomorphic

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Statement

Suppose G is a group and H is a Retract (?) of G. In other words, H has a Normal complement (?) in G: there exists a normal subgroup N of G such that NH = G and N \cap H is trivial.

Then, there may exist another normal complement M to H in G such that M is not isomorphic to N.

Proof

Example of the dihedral group

Further information: dihedral group:D8

Consider the following example:

  • G is the dihedral group of order eight:

\langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle.

  • H is the subgroup \{ x, e \}, i.e., a two-element cyclic subgroup.
  • There are two normal complements to H in G. The subgroup N = \{ x, x^2, x^3, e \} is cyclic of order four and the subgroup M = \{ e, x^2, ax, a^3x \} is a Klein-four group.