# A conjecture about non-nilpotent groups

A finite group, $$G$$, is nilpotent if its upper central series terminates (at $$i \in \mathbb{N})$$ with $$Z^i(G)=G$$, where $$Z^i(G)$$ is it's $$i$$-th center which can be described as $$\{x \in G\mid \forall y \in G : [x,y]\in Z_{i-1}(G)\}$$ where $$[x,y]$$ is the commutator of $$x$$ and $$y$$ in $$G$$ (in other words, it is the set of elements that commute with one another up to an element of the $$(i-1)$$-th center).

A finite group, $$G$$, is non-nilpotent if it is not nilpotent. In this case, we still have, by finiteness of $$G$$ that the upper central series will terminate after finitely many steps, but now we have an $$i \in \mathbb{N}$$ at where $$Z^i(G)=Z^{i+1}(G)$$, and yet $$Z^i(G)\neq G$$, but rather a proper subgroup of $$G$$. In this case, since the upper centers are all characteristic and thus normal, we can consider the quotient group, $$Q=G/Z^i(G)$$. This group must be centerless, else we could take the pullback of $$Z(Q)$$ under the natural projection to obtain $$Z^{i+1}(G)\supset Z^i(G)$$ with equality not holding, contradicting our assumption that the upper central series stabilized at $$i$$. This gives a rather elegant characterization of non-nilpotent finite groups:

A non-nilpotent finite group $$G$$ can be viewed as an extension of a nilpotent normal subgroup, $$U=Z^i(G)$$ by a centerless quotient group, $$Q$$. In the most extreme case, the central series terminates immediately in the trivial group, and you obtain a centerless $$G=Q$$. This shows that the centerless groups are the primitive elements of the class of non-nilpotent finite groups with the rest being constructed from them by extensions of nilpotent groups by centerless groups.

A (finite) group, $$G$$, is complete if it is centerless and isomorphic to its automorphism group $$G\simeq\text{Aut}(G)$$.

Conjecture: Suppose we have a non-nilpotent finite group $$G$$, whose upper central series terminates at $$i>0$$, meaning that $$G$$ is not centerless, and $$Z^i(G)=Z^{i+1}(G)$$. Let $$U=Z^i(G)$$, and suppose further that the centerless quotient $$Q=G/U$$ is complete, meaning $$Q\simeq\text{Aut}(Q)$$. Lastly, suppose that the upper central series for $$\text{Aut}(G)$$ terminates at some $$i_1\geq i$$, that is, the upper central series for $$\text{Aut}(G)$$ is at least as long as the upper central series for $$G$$. Note in particularly that $$\text{Aut}(G)$$ is non-nilpotent, and so has a centerless quotient $$Q_1$$. Then the claim is that the centerless quotients are isomorphic, $$Q_1\simeq Q$$.

I came up with the conjecture based on computer searches in GAP looking at the automorphism series for various small finite groups as far as I could construct them in GAP (many eventually reach very large groups and run into issues with memory or the size of the computation). Looking at the upper central series for each term of the automorphism series and comparing how they changed as you step through the automorphism series lead me to propose that the above conjecture holds in general for finite groups. Naturally a counterexample will be enough to disprove the claim, but I didn't find any looking at over 50 different automorphism series, so it may not be easy to find if it exists. I'm not really sure how to approach trying to prove this either. I don't really know enough about the relationship between a group and its automorphism group to leverage the givens in this conjecture productively.

Edit 1: This is a response to David A. Craven's answer: I took the liberty of constructing the proposed group in GAP for $$n=7$$. To do this, I needed to put all three of the $$G_i$$'s into permutation groups so that the direct product operation in GAP would also return a permutation group, so that fast algorithms would be available for subsequent calculations. Specifically I used:

G1:=SymmetricGroup(7);
G2:=Group((1,2,3,4),(1,2)(3,4)); #This is D8
G3:=Group((1,2,3),(4,5,6),(7,8,9)); #This (Z_3)^3
#That G2 and G3 are as claimed can be checked by IdSmallGroup, is should return (8,3) for G2 and (27,5) for G3.
#Now we can call DirectProduct and the result will be a permutation group
G:=DirectProduct(G1,G2,G3);
#Because this is a permutation group, computation of the upper central series is 'easy' for GAP, trying this without doing the above preparations lead to a computation that was taking minutes and still hadn't finished
U:=UpperCentralSeriesOfGroup(G);


From the above I obtained that $$G$$ has upper central series of length $$2$$. With $$Z(G)\simeq (54,15)$$, and $$Z^2(G)\simeq (216,151)$$ (Where (x,y) is the GAP library id for order x, id y.). GAP StructureDescription for U[1] confirms that $$Z^2(G)\simeq\mathbb{Z}_3^3\times D_8$$. Next I constructed the quotient and confirmed it was the symmetric group (necessarily $$S_7$$ by an easy order argument). After a computation that took a few seconds I got the automorphism group, $$A1$$, but it was not in the form of a permutation group, so I used NiceMonomorphism and then NiceObject to obtain a permutation group isomorphic to the automorphism group. This allows for fast(er) algorithms to used for computing its upper central series, which I found to be of length $$2$$ with $$Z(A1)\simeq (8,5)$$ and $$Z^2(A1)\simeq (32,46)$$ (again using GAP ids for the groups). The size of $$A1$$ is 905748480 which is twice that predicted in the answer (Which would be 452874240 if it had the structure given in the answer). I was able to get GAP to give me a structure description for $$A1$$, which was $$\mathbb{Z}_2^2\times D_8\times S_7\times \text{PSL}(3,3)$$. I suspect the comment that the proposed automorphism group is, in fact, a subgroup of index 2 in the full automorphism group is correct. The quotient $$Q1$$ is of order 28304640 (and so obviously not isomorphic to $$S_7$$), GAP gives a structure description of $$S_7\times \text{PSL}(3,3)$$ for this quotient. (So it is still a counter example for $$i_1=i$$ case).

• During the computer search, how many non-nilpotent groups $G$ could you find for which $i(G)\leq i(\operatorname{Aut}(G))$ and $i(G)\geq1$? Nov 2, 2021 at 16:57
• I did not keep an explicit count. I was mostly looking for patterns, and this was one of them that stood out. I don't think I encountered many cases where $i(\text{Aut}(G))<i(G)$ until all of the upper centers were elementary abelian (another conjecture I plan to ask about later). It's quite common in the earlier steps of the automorphism series to see the behavior you're asking about. Groups of orders 16, 24, and 32 provide many starting examples to look at for this behavior (orders<15 generally stabilize too quickly to reveal much in the way of interesting behavior). Nov 2, 2021 at 20:17

I ran a brute-force search on groups of small order using the following code:

findCounterexamples := function( range )
local Ids, id, G, A, UCS, UCS1, i, i1, Q, AQ, Q1;
Ids := IdsOfAllSmallGroups( range, IsNilpotent, false );;
for id in Ids do
G := SmallGroup( id[1], id[2] );
A := AutomorphismGroup( G );
UCS := UpperCentralSeries( G );
UCS1 := UpperCentralSeries( A );
i := Length( UCS );
i1 := Length( UCS1 );
if i1 < i then
# Aut(G)'s central series is shorter than G's
continue;
fi;
Q := G/UCS[1];
AQ := AutomorphismGroup( Q );
if IsomorphismGroups( Q, AQ ) = fail then
# Q is not complete
continue;
fi;
Q1 := A/UCS1[1];
if IsomorphismGroups( Q, Q1 ) = fail then
Print( "SmallGroup(", id[1], ",", id[2], ")\n" );
Print( "G = ", StructureDescription( G ), "\n" );
Print( "G has UCS of length ", Length( UCS ), "\n");
Print( "Q = ", StructureDescription( Q ), "\n");
Print( "Aut(G) = ", StructureDescription( A ), "\n" );
Print( "Aut(G) has UCS of length ", Length( UCS1 ), "\n");
Print( "Q1 = ", StructureDescription( Q1 ), "\n\n");
fi;
od;
end;


Letting it run over the groups of order $$\leq 72$$ produced three counterexamples:

SmallGroup(48,42)
G = C2 x C2 x (C3 : C4)
G has UCS of length 2
Q = S3
Aut(G) = S3 x ((((C2 x C2 x C2) : (C2 x C2)) : C3) : C2)
Aut(G) has UCS of length 2
Q1 = S3 x (((C2 x C2 x C2 x C2) : C3) : C2)

SmallGroup(54,12)
G = C3 x C3 x S3
G has UCS of length 2
Q = S3
Aut(G) = S3 x GL(2,3)
Aut(G) has UCS of length 2
Q1 = S3 x S4

SmallGroup(72,48)
G = C2 x C6 x S3
G has UCS of length 2
Q = S3
Aut(G) = C2 x S3 x S4
Aut(G) has UCS of length 2
Q1 = S4 x S3


EDIT: There also exist counterexamples with $$i_1 > i$$, e.g.

SmallGroup(96,78)
G = C4 x C4 x S3
G has UCS of length 2
Q = S3
Aut(G) = ((C2 x C2 x ((C2 x C2 x C2 x C2) : C3)) : C2) x S3
Aut(G) has UCS of length 3
Q1 = (((C2 x C2 x C2 x C2) : C3) : C2) x S3

SmallGroup(192,1030)
G = C2 x C4 x C4 x S3
G has UCS of length 2
Q = S3
Aut(G) = ((((C2 x C2 x ((C2 x C2 x ((C2 x C2 x C2 x C2) : C2)) : C2)) : C3) : C2) : C2) x S3
Aut(G) has UCS of length 3
Q1 = S3 x ((((C2 x C2 x (((C2 x C2 x C2 x C2) : C2) : C2)) : C3) : C2) : C2)

• I wonder if we instead make the inequality strict ($i_1>i$) if there are still any counterexamples. Nov 3, 2021 at 12:35
• There are indeed, I've edited the answer to include some. Nov 3, 2021 at 13:53
• I noticed that in all of these cases $Q_1\simeq Q\times P$ for some other (centerless) group $P$ (which I suspect is also complete). I wonder if this is always that case when the original conjecture fails... Nov 3, 2021 at 14:48

Easy counterexamples appear to be as follows: let $$G_1$$ be a centreless complete group, say $$S_n$$ for $$n\geq 7$$. Let $$G_2$$ be a non-abelian $$p$$-group isomorphic to its own automorphism group, say $$D_8$$. Let $$G_3$$ be an abelian group with a big automorphism group, say an elementary abelian one, $$C_3^3$$. Let $$G=G_1\times G_2\times G_3$$. The quotient by the hypercentre is just $$G_1\cong S_n$$. The automorphism group of $$G$$ has a subgroup of index $$2$$ the direct product of the automorphism groups, so is $$S_n\times D_8\times GL_3(3)$$. The hypercentre is $$D_8\times C_2$$, which has the same nilpotency class as the original hypercentre, and the quotient has $$S_n\times PGL_3(3)$$ of index $$2$$.

• Interesting...as far as I know, $D_8$ is the only example of a (finite) $p$-group isomorphic to its own automorphism group (I have a complete list of groups isomorphic to their own automorphism group for groups of order$<512$ and $D_8$ is the only $p$-group in that list). Nov 3, 2021 at 12:31
• I think it's an open problem (might be in Kourovka notebook) whether $G = D_8$ is the only non-trivial $p$-group such that $Aut(G) \cong G$.
– spin
Nov 3, 2021 at 13:37
• Also, I guess your example works, but the automorphism group is not quite right. I think $S_n \times D_8 \times GL_3(3)$ has index $2$ in $Aut(G)$. This is because $Hom(S_n,Z(D_8))$ is nontrivial, see here. Maybe $G_1 = A_n$ is easier (or some other centerless complete group with no subgroup of index $2$)?
– spin
Nov 3, 2021 at 13:53
• @spin Yes, you are right. OK, it contains this. Nov 3, 2021 at 14:36
• Few more comments: I think the subgroup in the question is called the hypercenter, i.e. the stable term in the upper central series. This is in general smaller than the Fitting subgroup, but in your example it seems to be the same. Also shouldn't the Fitting subgroup be $D_8 \times C_2$ and quotient $PGL_3(3)$?
– spin
Nov 4, 2021 at 0:51