# Nilpotent group with ascending and descending central series different?

This may turn out to be a bit embarassing, but here it is. It is probably not true that the ascending and descending central series (*) of a nilpotent group have the same terms. (Or at least one of MacLane-Birkhoff, Rotman and Jacobson would have mentioned it.) However, I have been unable to find an example where they are different. I thought I had a sketch of proof that they are always equal, but there is a gap, of the kind where you feel it is not patchable.

I've proved it for a few nilpotent groups (dihedral of the square, any group of order p^3, the Heisenberg groups of dimension 3 and 4 over any ring -- I think the argument extends to any dimension), and checked a few more exotic examples in the excellent Group Properties Wiki. So,

What is the simplest (preferably finite) nilpotent group such that its a.c.s. and d.c.s. are different?

and

Do the a.c.s. and d.c.s. coincide in some interesting general case?

(*) For completeness, the ascending central series of a group G is defined by Z_0 = 1, Z_{i+1} = the pullback of Z(G/Z_i(G)) along the projection, and the descending central series by G_0 = G, G_{i+1} = [G,G_i]. The group G is nilpotent iff ever Z_m = G or G_m = 1. It turns out that if such m exists it is the same for both.

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In your notation for the lower/descending central series, you have $G_1=[G,G]$; this is not the standard notation. Usually, $G_2=[G,G]$, and the group is nilpotent of class $c$ if $G_{c+1}=1$. – Arturo Magidin Jul 6 '10 at 18:01
I've seen both conventions used in different texts, but I'm certainly no group theorist and I believe you. I do like to assign index 0 to "places where we haven't done anything yet" though... – Pietro KC Jul 6 '10 at 19:40
The benefit of Arturo's suggested notation is that $[G_i,G_j]\subset G_{i+j}$. – Tom Church Jul 7 '10 at 1:46
Arturo's notation is standard. I doubt any serious modern text is doing it differentially. You might be confusing the lower central series which starts at 1 and the derived series which starts at 0. – Yiftach Barnea Jul 7 '10 at 6:50
MacLane-Birkhoff do it as I did above, and for my modest immediate needs the index 0 rationale seemed more intuitive. Anyway, thanks for the heads-up! – Pietro KC Jul 8 '10 at 4:56

It's false even if $m=2$. Try the product of a group of order $2$ and a dihedral group of order $8$. The center has order $4$ and the quotient by the center is abelian. The commutator subgroup has order $2$.

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Also, there are no strictly smaller ones, as neither of the nonabelian groups of order 8 works, and any other nilpotent group of order <16 is abelian. – Harry Altman Jul 6 '10 at 10:39
Thanks, this is simpler than I expected. I had this funny notion in the back of my head that the property "a.c.s. equal to d.c.s." would be preserved by direct products, I'm not sure why... – Pietro KC Jul 6 '10 at 19:46
Probably because it is true that $Z_i(G\times H)=Z_i(G)\times Z_i(H)$ and $(G\times H)_k = G_k\times H_k$ (the former also holds in infinite direct products, but the latter need not). It is true that if the lower and upper central series of $G$ coincide, and the same holds for $H$, and $G$ and $H$ have the same nilpotency class, then the upper and lower central series of $G\times H$ will coincide as well. But if the nilpootency classes differ, then you'll have the same situation as with Tom's example (if $class(H)\lt c=class(G)$, look at $Z_{c-1}(G\times H)$ vs. $(G\times H)_2$). – Arturo Magidin Jul 9 '10 at 17:03

Nobody seems to have answered the second query so far; one important class of groups in which the two series coincide is the class of $p$-groups of maximal class. A $p$-group $G$ of order $p^n$ is of maximal class if the nilpotency class of $G$ is $n-1$. The nonabelian groups of order $p^3$ are an example.

If $G$ is a $p$-group of maximal class, then the lower and upper central series coincide; this is essentially for the same reason as they do in the case of groups of order $p^3$, as Noah Snyder comments: there just isn't enough room for the series to differ.

If $G$ is of order $p^n$ and class $n-1$, then letting $G_2=[G,G]$ and $G_{i+1}=[G_i,G]$ (this is different from the way it is defined in the question as I write this; here, the group has class exactly $c$ if and only if $G_c\neq 1$ and $G_{c+1}=1$), we have $|G_i:G_{i+1}|=p$ for $i=2,3,\ldots,n-1$; similarly, $|Z_{j+1}(G):Z_j(G)| = p$ for $j=0,1,\ldots,n-2$. Since $G_{n-1}\subseteq Z_{1}(G)$ and both are of order $p$, they are equal; taking the quotient gives a group of maximal class and order $p^{n-1}$, and an inductive argument gives the equality among the rest of the terms.

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Thanks for your answer, I wish I could accept both yours and Tom's. That's a neat way to organize the facts; the proof sketch I referred to above is pretty much the one you have here, but I didn't think to restrict to p-groups of maximal class. – Pietro KC Jul 6 '10 at 19:53
Note that if $G$ is a nilpotent group of class $c$, then, following my notation, you always have $G_{c-k}\subseteq Z_{k+1}(G)$. To see this, proceed by induction on~$k$; if $k=0$, then $1=G_{c+1}=[G_c,G]$, so $G_c\subseteq Z_1(G)$. Assuming that $G_{c-k}\subseteq Z_{k+1}(G)$, then it follows that $G_{c-k+1}Z_{k+1}/Z_{k+1}\subseteq Z_{k+2}(G)/Z_{k+1}(G)$ (since $[G_{c-k+1}Z_{k+1}/Z_{k+1},G/Z_{k+1}]=1$ by the assumption), thus $G_{c-k+1}\subseteq Z_{k+2}$, as needed. In the case of maximal class (and in other instances), you have equalities all the way through. – Arturo Magidin Jul 7 '10 at 1:02

There's a good reason that the series have to be the same for groups of order p^3. Namely, the quotient G/Z(G) (in general) and the quotient G/[G,G] (for non-cyclic nilpotent groups) are never cyclic. The former result is quite easy, while the latter follows from the Burnside basis theorem. So in a group of order p^3 there's just not enough room for Z(G) and [G,G] to be different.

But in a group of order p^4 there's no such obstruction. For a group of order p^4 there's enough room for Z(G) and [G,G] to be distinct and have orders p and p^2. In Tom's example the commutator subgroup is the smaller one. But you could also try to do it the other way around. So we want Z(G) to have size p, and G/Z(G) to be a non-abelian group of order p^3 (if it were abelian then Z(G)=[G,G]). Anyone know what H^2 of the groups of order p^3 are?

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Not only can the two central series be different, there are many qualitative properties that the descending central series (also called the lower central series) satisfies that the ascending central series (also called the upper central series) usually does not. For instance, the lower central series is strongly central (as pointed out by Tom Church in his comment, though he didn't use that word) but the upper central series (turned around) need not be. At least half the members of the lower central series are abelian, but this need not be true of the upper central series. All the members of the lower central series are verbal subgroups, but the members of the upper central series need not even be fully invariant.

http://groupprops.subwiki.org/wiki/Nilpotent_not_implies_UL-equivalent

[DISCLOSURE: I wrote the material on that page and in most of the linked-to pages]

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Can we find an example of a central series that coincides not with the ascending nor with the descending one? Does $Z_2 \times \text{Dih}_8 \times \text{Dih}_{16}$ serve the purpose? – Student Oct 31 '14 at 13:55

Other examples of natural families of $p$-groups where the a.c.s. coincide with the d.c.s. are the following:

Let $R=\mathbb{Z}/(p^n)$ or $R=\mathbb{F}_p[t]/(t^n)$.

Let $G_{k}$ be the $k$-congruence subgroup of $SL_d(R)$, i.e. the matrices which form the kernel of mod $p^k$ or mod $t^k$ respectively.

Let $G=G_1$. Then $G$ is a $p$-group and if $p \nmid d$, then the congruence series coincides with the a.c.s. and the d.c.s.. The easiest way to see this is to use the associated graded Lie algebra, but I will omit the details as that will be too long.

These examples are coming from quotients of pro-$p$ groups. There are clearly generalizations to quotients of $\Lambda$-simple pro-$p$ groups. This might also work also for quotients of the Nottingham group (but I am not 100% sure).

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