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Today in my research, I had to use fairly explicitly the rather tautological property of finite cyclic groups that every normal subgroup is characteristic, i.e. fixed by all automorphisms. This got me wondering:

do (finite) groups with the property that every normal subgroup is characteristic have a name and/or can they be completely classified? Generally, has this property been investigated at all?

Apart from cyclic groups, some groups possessing the above property that immediately come to mind are simple groups, symmetric groups, semi-dihedral groups, and dihedral groups of twice odd order (however not of twice even order).

This is as far as I got on my short walk home (apart from some false claims, see comments). I suspect that this property might be well studied.

Edit: The reference that Beren Sanders provided in his answer and the references to and from it all deal with $p$-groups. I still haven't been able to find anything about arbitrary finite groups. Some of the questions that $p$-group theorists ask are just not terribly interesting in the case of arbitrary groups. E.g. the paper that Beren Sanders mentioned proves that every finite $p$-group is contained in another finite $p$-group in which every normal subgroup is characteristic. The same statement for arbitrary finite groups is trivial: just embed your group into a symmetric group. I would still be surprised if nobody had tried to say something reasonably general about finite groups with this property.

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  • $\begingroup$ Sorry I can't help, but, if it's not too much trouble, can someone give an example of a normal subgroup that isn't characteristic? $\endgroup$
    – Gerry Myerson
    Commented Jan 1, 2011 at 15:44
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    $\begingroup$ Any subgroup of index 2 in the quaternion group of order 8. $\endgroup$
    – Torsten Ekedahl
    Commented Jan 1, 2011 at 15:48
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    $\begingroup$ @Alex: are you sure that quotient groups inherit the property? It seems that the group $G:=S_5 \times S_7$ has abelianization that is a product of two cyclic groups of order 2. In particular, every non-trivial subgroup of the abelianization is normal, but not characteristic. On the other hand, if I am not mistaken, the only normal subgroups of $G$ are also characteristic. $\endgroup$
    – damiano
    Commented Jan 1, 2011 at 16:49
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    $\begingroup$ Actually, here is a counterexample to the original claim that the property is inherited by normal subgroups: $V_4$ in $S_4$. One can probably obtain many more by taking semi-direct products of groups that don't have the property by their outer automorphism groups. $\endgroup$
    – Alex B.
    Commented Jan 1, 2011 at 16:51
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    $\begingroup$ Normal subgroups of complete groups are characteristic. $\endgroup$
    – Someone
    Commented Jan 11, 2011 at 10:11

2 Answers 2

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The question in which $p$-groups all normal subgroups are characteristic is a fairly old problem in the theory of finite $p$-groups. It seems difficult to assess because of the fact that neither subgroups nor factors of a group which has none but characteristic normal subgroups need retain this property.

These are the first two sentences of the following 2009 paper published in the Israel Journal of Mathematics:

B. Wilkens. $p$-groups without noncharacteristic normal subgroups. Isr. J. Math. 172, 357–369 (2009). Zbl 1188.20014

As a non-expert I have nothing more to add.

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  • $\begingroup$ That's an interesting pointer, thanks! $\endgroup$
    – Alex B.
    Commented Jan 2, 2011 at 2:39
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Here is a small contribution. If no two factors in a composition series for $G$ are isomorphic, then every normal subgroup of $G$ is characteristic. In fact, in groups of this type, no two distinct normal subgroups can be isomorphic. To see this, suppose $M$ and $N$ are distinct and isomorphic. Look at a composition series obtained by refining the series $1 \subseteq M \cap N < M <MN \subseteq G$. Let $S$ be a simple group occurring as a composition factor between $M \cap N$ and $M$. Then $S$ is not isomorphic to a composition factor of $M \cap N$, but it is a composition factor of $M$ and hence also of the isomorphic group $N$. Since $S$ is a composition factor of $N$ but not of $M \cap N$, it is a composition factor of $N/(N \cap M)$. Since this group is isomorphic to $MN/M$, we see that $MN/M$ has $S$ as a composition factor. This is a contradiction because then $G$ has two composition factors isomorphic to $S$: one between $M \cap N$ and $M$ and another between $M$ and $MN$.

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  • $\begingroup$ (1) I'd like to recall the old problem of study of the $p$-groups all of whose maximal subgroups are characteristic. (2) Does there exist a noncyclic group of exponent $p$, all of whose maximal subgroups are characteristic? (3) Is it true that if all subgroups of order $p$ from $\text{Z}(G)$ are characteristic in the $p$-group $G$, then $|\Omega_1(\text{Z}(G))|$ is bounded? $\endgroup$
    – yakov
    Commented Jun 29, 2016 at 9:45

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