Complements to normal subgroup need not be automorphic

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Revision as of 18:39, 8 November 2008 by Vipul (talk | contribs) (New page: ==Statement== Suppose <math>G</math> is a group, <math>N</math> is a normal subgroup, and <math>H</math> and <math>K</math> are permutable complements to <math>N</math> in <math>G</ma...)
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Statement

Suppose G is a group, N is a normal subgroup, and H and K are permutable complements to N in G. Then, it is not necessary that there exists an automorphism of G sending H to K.

Related facts

Proof

A generic example

Let A be any non-Abelian group. Consider G = A \times A and the subgroup N = A \times \{ e \}. Let H be the subgroup \{ e \} \times A and K be the subgroup \{ (a,a) \mid a \in A \}.

Note that:

  • N is normal in G: In fact, it is a direct factor of G.
  • H is a permutable complement to N in G.
  • K is a permutable complement to N in G.
  • H is normal in G: In fact, it is a direct factor of G.
  • K is not normal in G: Pick a,b \in A such that a,b do not commute. Then, we have (b,e)(a,a)(b,e)^{-1}) = (bab^{-1},a). Thus, a conjugate of an element in K lies outside K.