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