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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.

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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    $\begingroup$ As Hendrik Lenstra once said to me, there is also a very beautiful classification of groups that have exactly one non-normal subgroup.... :) $\endgroup$ May 20, 2010 at 3:01
  • $\begingroup$ :) :) $\endgroup$ May 20, 2010 at 3:02
  • $\begingroup$ I like this question and all comments! $\endgroup$
    – Kerry
    May 20, 2010 at 3:21
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    $\begingroup$ @Bjorn: is that just a joke, or is there really a classification of groups having exactly one conjugacy class of non-normal subgroups? $\endgroup$ Nov 15, 2010 at 14:47
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    $\begingroup$ In fact, Otto Schmidt (1926) has classified the finite groups all of whose nonnormal subgroups are conjugate. Later, in 1938, he also has classified the finite groups with exactly two classes of nonnormal subgroups. $\endgroup$
    – yakov
    Jun 29, 2016 at 9:55

4 Answers 4

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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 = (ax1)(ax2)(ax3) = ax1*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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  • $\begingroup$ Ahh, looks like I was a bit hasty. Thanks for the great answer! $\endgroup$ May 20, 2010 at 3:42
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    $\begingroup$ For people who stumble upon this (recently bumped) post, it might be unclear what you mean by "your example" and "your group". Perhaps link to the original revision of the question could be added. Another thing which is a bit unfortunate is that in several places the SE software interprets * as a sign for italics, although you probably want to have there $(a*x_1)*(a*x_2)*(a*x_3)$, etc. $\endgroup$ Jan 2, 2020 at 13:57
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The keyword is "Hamiltonian group". There is a classification of them, modulo the classification of abelian periodic groups (that is, torsion groups) with elements of odd order. You'll find the details, due to Dedekind and Baer, in W. R. Scott's Group Theory, for example.

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  • $\begingroup$ 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. $+^+$ $\endgroup$
    – Mikasa
    Apr 6, 2013 at 7:49
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Yes, these are called Hamiltonian groups.

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If we replace “(say finite)” by “(say connected Lie)”, then the fact that such groups are abelian may be first observed in:

Ahrens, W., Über Transformationsgruppen, deren sämtliche Untergruppen invariant sind, Hamb. Mitt. 4, 72-78 (1902). ZBL33.0162.02.

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