13
$\begingroup$

Let $G$ be a group such that, for every subgroup $H$ of $G$, there exists a normal subgroup $K$ of $G$, such that $H$ is isomorphic to $K$. Under such conditions, can we determine the structure of $G$ ?

This question comes from the discussion of Dedekind group in group theory, which is one such that all of its subgroups are normal. We know that a Dedekind group is an Abelian group or a direct product of the quaternion group $Q_8$ and an Abelian group $A$, where $A$ has no elements with order $4$.

So my problem can be regarded as a promotion of Dedekind group that only requires isomorphism. I don't know how to deal with this case.

$\endgroup$
10
  • 17
    $\begingroup$ For a finite group, it implies that every Sylow subgroup is normal, and hence this is a nilpotent group, reducing the study to finite $p$-groups. $\endgroup$
    – YCor
    Aug 25, 2020 at 6:12
  • 3
    $\begingroup$ @vrz : Automorphisms of a group $G$ send normal subgroups to normal subgroups, and non-normal subgroups to non-normal subgroups. Hence every subgroup of $G$ is mapped to a normal subgroup via some automorphism if and only if every subgroup of $G$ is normal (so $G$ is Dedekind). $\endgroup$ Aug 25, 2020 at 10:37
  • 2
    $\begingroup$ The Heisenberg group over $\mathbb{Z}$ (en.wikipedia.org/wiki/Heisenberg_group) is an infinite nilpotent group which is not Dedekind but satisfies your requirements. (Maybe some finite quotient satisfies them too.) $\endgroup$
    – Luc Guyot
    Aug 25, 2020 at 11:24
  • 2
    $\begingroup$ @LucGuyot Is this easy to see? Sorry but straight away I don't even see whether there are any non-cyclic non-normal subgroups. $\endgroup$ Aug 25, 2020 at 12:14
  • 2
    $\begingroup$ For $2$-groups the first failures are given by $D_{16}$, $SD_{16}$ and another group of order $16$ (no 3 in the SmallGroup library). The first 2 groups contains a $C_2\times C_2$ subgroup (but no normal ones) and the second has a $C_4$-sunbgroup (but no normal ones). For order $32$ the property fails for $24$ out of $51$ groups and for order $64$ it fails for $205$ out of $267$ groups. $\endgroup$ Aug 25, 2020 at 13:38

1 Answer 1

12
$\begingroup$

Let us call $G$ a generalised Dedekind group if every subgroup of $G$ is isomorphic to a normal subgroup of $G$.

As expected, Dedekind groups are generalised Dedekind groups. In addition, YCor has established that a finite generalised Dedekind group $G$ is nilpotent, since its $p$-Sylow subgroups are necessarily normal.

Note that the fundamental group of the Klein bottle, that is, $\mathbb{Z} \rtimes_{-1} \mathbb{Z} = \left\langle a, b \, \vert \, aba^{-1} = b^{-1} \right\rangle$ is an infinite generalised Dedekind group which is not nilpotent.

Here is a family of finite nilpotent groups which are generalised Dedekind groups but not Dedekind groups.

Claim. Let $p$ be a prime number and let $u$ be an integer such that $u \equiv 1 \text{ mod } p \mathbb{Z}$ and $u^p \equiv 1 \text{ mod } p^2 \mathbb{Z}$. Let $$G(p, u) \Doteq \mathbb{Z}/p^2 \mathbb{Z} \rtimes_u \mathbb{Z}/p \mathbb{Z}$$ where the conjugation by $a \Doteq (0, 1 + p \mathbb{Z})$ induces the multiplication by $u$ on $\mathbb{Z}/p^2 \mathbb{Z}$, i.e., $aba^{-1} = b^u$ where $b \Doteq (1 + p^2 \mathbb{Z}, 0)$. Then $G(p, u)$ is a generalised Dedekind group. If in addition $u \not\equiv 1 \text{ mod } p^2 \mathbb{Z}$, then $G(p, u)$ is not Dedekind.

Proof. Let $H$ be a subgroup of $G(p, u)$. It is easy to check that $H$ can be generated by two elements $(x, y) = (b^k, b^la^m)$ with $k \in \{0, 1, p\}$ and $l, m \in \mathbb{Z}$. Let us show first that $G(p, u)$ is a generalised Dedekind group.

If $k = 0$ or $m \equiv 0 \text{ mod } p\mathbb{Z}$ , then $H$ is a cyclic subgroup of order at most $p^2$ and hence isomorphic to either $\{1\}$, $\langle b \rangle$ or $\langle b^p \rangle$ which are normal in $G(p, u)$. Indeed, we have $(b^la^m)^p = b^{l(1 + u^m + \dotsb + u^{m(p-1)})}$ with $1 + u^m + \dotsb + u^{m(p-1)} \equiv p \bmod{p\mathbb{Z}}$, hence the order of $b^la^m$ is at most $p^2$.

If $k = 1$, then $H = \langle b, a^m \rangle$ which is easily seen to be normal in $G(p, u)$.

Let us assume eventually that $k = p$ and $m \not\equiv 0 \text{ mod } p\mathbb{Z}$. Then we have $H = \langle b^p, b^l a^m \rangle$. If $(b^l a^m)^p \neq 1$, then $b^l a^m$ generates $H$ so that $H$ is isomorphic to a cyclic normal subgroup of $G(p, u)$. Otherwise $\langle b^p \rangle \cap \langle b^l a^m \rangle = \{1\}$. Since $\langle b^p \rangle$ is central in $G(p, u)$, the subgroup $H$ is isomorphic to the Abelian normal subgroup $\langle b^p, a \rangle \triangleleft G(p, u)$.

Thus we have established that $G(p, u)$ is a generalised Dedekind group. We complete the proof by observing that $G(p, u)$ is Abelian if and only if $u \equiv 1 \text{ mod } p^2 \mathbb{Z}$.

Taking $p = 2$ and $u = 3$ yields the dihedral group of order $8$ which is the smallest finite generalised Dedekind ring which is not Dedekind. Taking $p = 3$ and $u = 4$ provides us with an example of a $3$-group which is a generalised Dedekind but not Dedekind.


Addendum. Geoff Robinson and YCor have observed in the comments attached to the question that, more generally, the groups of order $p^3$, for $p$ a prime number, are generalised Dedekind groups. If $p \neq 2$, there are only $2$ isomorphism classes of non-Abelian groups of order $p^3$: the class of $G(p, u)$ (which does not depend on $u$) and the class of the Heisenberg group modulo $p$, that is, $H(\mathbb{Z}/p\mathbb{Z}) \Doteq (\mathbb{Z}/p\mathbb{Z})^2 \rtimes_A \mathbb{Z}/p\mathbb{Z}$ with $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The proof becomes obvious once we have observed that the centers of these two groups are cyclic of order $p$ and that groups of order $p^2$ are Abelian. If $p = 2$, the latter two classes collapse in one, the isomorphism class of the dihedral group of order $8$, and a new class must be considered, namely the isomorphism class of the quaternion group.

$\endgroup$
1
  • 1
    $\begingroup$ An interesting (to me at any rate) example of a $p$-group $P$ (for $p$ odd) which is not generalized Dedekind, yet has an Abelian subgroup of index $p$ is the wreath product $\left( \mathbb{Z}/p\mathbb{Z} \right) \wr \left( \mathbb{Z}/p\mathbb{Z} \right) $. (since $P$ has a cyclic subgroup of order $p^{2}$, but no normal cyclic subgroup of order $p^{2}$). $\endgroup$ Aug 26, 2020 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.