Conjugacy-closedness is not join-closed

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 $a^{2}x$ in this subgroup are conjugate in $G$.