# Abelian normal is not join-closed

## Statement

It is possible to have a group $G$ with Abelian normal subgroups $H, K$ such that the join $\langle H, K \rangle$ is not an Abelian normal subgroup.

## Related facts

A group obtained as a join of Abelian normal subgroups is termed a group generated by Abelian normal subgroups. Such groups have a number of nice properties.

## Proof

### Example of the dihedral group

{{further|[[Particular example::dihedral group:D8]}}

Let $G$ be the dihedral group of order eight:

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

Let $H, K$ be subgroups of $G$ given as follows:

$H = \langle a \rangle, \qquad, K = \langle a^2, x \rangle$.

Both $H$ and $K$ are Abelian normal subgroups, but the join of $H$ and $K$, which is the whole group $G$, is not an Abelian normal subgroup.

### Example of the quaternion group

Further information: quaternion group

In the quaternion group, the cyclic subgroups generated by $i$ and $j$ are both Abelian normal of order four, but their join, which is the whole group, is not Abelian.

### Any non-Abelian group of prime-cubed order

Further information: prime-cube order group:p2byp, prime-cube order group:U3p

If $p$ is an odd prime, the two non-Abelian $p$-groups of order $p^3$ again offer examples of groups with Abelian normal subgroups whose join is not normal. In both cases, there are multiple subgroups of order $p^2$, that are Abelian and normal, and whose join is the whole group, which is not normal.