# Groups with all subgroups normal

Is there any sort of classification of (say finite) groups with the property that every subgroup is normal?

Of course, any abelian group has this property, but the quaternions show commutativity isn't necessary. More generally, the group

$\rlap{////////////////////////////////////////////////}\langle a, x_1,\ldots,x_n|\text{ } ax_i=x_ia, \text{ }a^2=1, \text{ }x_i^2=a, \text{ }x_i^{-1}x_jx_i=x_j^{-1}\rangle$

will have this property.(See answer below). If there isn't a classification, can we at least say the group must be of prime power order, or even a power of two?

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As Hendrik Lenstra once said to me, there is also a very beautiful classification of groups that have exactly one non-normal subgroup.... :) –  Bjorn Poonen May 20 '10 at 3:01
:) :) –  Bjorn Poonen May 20 '10 at 3:02
I like this question and all comments! –  Kerry May 20 '10 at 3:21
@Bjorn: is that just a joke, or is there really a classification of groups having exactly one conjugacy class of non-normal subgroups? –  Maurizio Monge Nov 15 '10 at 14:47

These are called Dedekind groups, and the non-abelian ones are called Hamiltonian groups. The finite ones were classified by Dedekind, and the classification extended to all groups by Baer. The non-abelian ones are a direct product of the quaternion group of order 8, an elementary abelian 2 group, and a periodic abelian group of odd order (or all of whose elements have odd order).

Periodic abelian groups all of whose elements have odd order can be quite complicated, but the finite ones are direct products of cyclic groups.

Your example does not have the property that all of its subgroups are normal when n ≥ 4. The subgroup generated by x1*x2*x3 is not normal, since (x1*x2*x3)^x4 = (a*x1)*(a*x2)*(a*x3) = a*x1*x2*x3, but x1*x2*x3 has order 2. For n = 3, your group is Q8 x 2, and so is Hamiltonian.

The cyclic group of order 6 and the direct product Q8 x 3 are two groups of non-(prime power) order with every subgroup normal.

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Ahh, looks like I was a bit hasty. Thanks for the great answer! –  Kevin Ventullo May 20 '10 at 3:42
Dear Mariano, may I ask you: For what sort of finite groups (which I am thinking about their structures and don't know them), there is at least one Hamiltonian subgroup? I mean, under which condition(s) for a group, one can find a Hamiltonian subgroup of it? Thanks and sorry for asking after a long time ago. $+^+$ –  Babak Sorouh Apr 6 '13 at 7:49