If $G$ is an abelian group, every subgroup $H$ of $G$ is normal.

I searched for non-abelian finite groups $G$ , such that every subgroup is normal and GAP showed only the groups $G'\times Q_8$ , where $Q_8$ is the quaternion-group of order $8$ and $G'$ is an abelian group with the additional property that $C_4$ is not a subgroup of $G'$. The largest group I found with GAP was $C_{15}\times Q_8$.

Is it true that every non-abelian finite group $G$ with the property that every subgroup of $G$ is normal, is isomorphic to $G'\times Q_8$ with an abelian group $G'$ without $C_4$ as a subgroup ?

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    $\begingroup$ Yes, these are called Hamiltonian groups and all have the form $Q_8\times A\times B$, where $A$ is an elementary abelian $2$-group, and $B$ is an abelian group of odd order. Here is a link. $\endgroup$ – James Feb 3 '16 at 22:54
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    $\begingroup$ See en.wikipedia.org/wiki/Dedekind_group $\endgroup$ – egreg Feb 3 '16 at 22:55
  • $\begingroup$ Thanks, James! So, $Q_8$ is the only group with the given property, if we do not consider direct products. $\endgroup$ – Peter Feb 3 '16 at 22:58
  • $\begingroup$ Are there infinite non-abelian groups with the given property ? $\endgroup$ – Peter Feb 3 '16 at 23:07

Hall's book "The Theory of Groups" proves (Theorem 12.5.4) that these groups, known as Hamiltonian groups, are all of the form $Q_8 \times A \times B$ where $A$ is an elementary $2$-group, $B$ is an abelian group where every element is of finite odd order. All groups of this form are Hamiltonian. Note there is no finiteness restriction on these $A$, $B$ groups, so there are also infinite group.

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