Conjugacy-closedness is not join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., conjugacy-closed subgroup) not satisfying a subgroup metaproperty (i.e., join-closed subgroup property).
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Statement

We can have a situation where H, K are conjuacy-closed subgroups of G but the join \langle H, K \rangle is not conjugacy-closed in G.

Proof

Example of the dihedral group

Further information: dihedral group:D8

Let G be the dihedral group of order eight:

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

Suppose H, K are subgroups given as follows:

H = \{ a^2, e \}, \qquad K = \{ x, e \}.

Observe that:

  • Since H and K are both subgroups of order two, they are both conjugacy-closed in G.
  • The join \langle H , K \rangle is an Abelian subgroup of order four. It is clearly not conjugacy-closed, because the elements x and a^2x in this subgroup are conjugate in G.