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Abelian normal is not join-closed


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.


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.