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I read that a direct product of a finite number of nilpotent groups is nilpotent. Here the definition of a nilpotent group is one that has a central series. A comment in my book following this claim says

If $G_{ij}$ is the $i^{th}$ term of a central series of the $j^{th}$ factor $H_j$, with $G_{ij}=G$ if the series has already terminated at $G$, then $\prod_j G_{ij}$ will be the $i^{th}$ term of a central series for $\prod_j H_j$.

My guess is that the central series for $\prod_j H_j$ is something like $$ 1\unlhd \prod_j G_{1j}\unlhd\prod_j G_{2j}\unlhd\cdots\unlhd\prod_j G_{rj}=\prod_j H_j $$ and additionally $$ \prod_j G_{i+1,j}/\prod_j G_{ij}\subseteq Z(\prod_j H_j/\prod_j G_{ij}). $$ I'm struggling to understand why the containment above is true. I think I need to show $$ \begin{align*} \prod_j g_{i+1,j}\prod_j G_{ij}\cdot\prod_j h_j\prod_j G_{ij} &= \prod_j g_{i+1,j}\prod_j h_j\prod_j G_{ij} \\ &= \prod_j h_j\prod_j g_{i+1,j}\prod_j G_{ij} \\ &= \prod_j h_j\prod_j G_{ij}\cdot\prod_j g_{i+1,j}\prod_j G_{ij} \end{align*} $$ but I just don't see why the second equality would be true. I'm sure there's a nice simple explanation, and I'd be glad to see it. Thanks.

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What I know is that for a group $G$, and its normal subgroup $H$, if both $H$ and $G/H$ are nilpotent, then so is $G$. I think this might be the case. – ShinyaSakai Jul 4 '11 at 8:34
It might be helpful to consider first the case of two groups and then proceed inductively. – Mark Jul 4 '11 at 8:41
@ShinyaSakai - what you say only holds for soluble groups, and not for nilpotent groups. – user1729 Jul 4 '11 at 9:00
@Shinya: $S_3$ is abelian-by-abelian, but not nilpotent. – Arturo Magidin Jul 4 '11 at 21:28
Indeed, the characterization in terms of Sylow subgroups shows that any finite nilpotent group of squarefree order must be cyclic. – Pete L. Clark Jul 4 '11 at 21:53
up vote 5 down vote accepted

Let $G=H\times K$. Note that $[(h_1, k_1), (h_2, k_2)] = ([g_1, g_2],[k_1, k_2])$, and so $[H\times K, H\times K]=[H, H] \times [K, K]=H_1\times K_1$. Similarly, $[H\times K,H_1\times K_1]= [H, H_1] \times [K, K_1]=H_2 \times K_2$, etc.

Assuming $H$ and $K$ are nilpotent, there exists and $i$ such that $G_i=H_i\times K_i=\langle 1\rangle$, and so $G$ is nilpotent.

The class of a nilpotent group $Q$ is defined to be the unique number $c$ such that $Q_c$ is trivial but $Q_{c-1}$ is non-trivial. Then, for $G=H\times K$ where $H$ is of class $c$ and $K$ is of class $d$, then the above working shows that $G$ is of class at most $\max(c, d)$.

Thus, the direct product of two nilpotent groups is nilpotent, and a simple inductive argument completes the proof.

This result is a specific case of a well-known result, called Fitting's Theorem, and the working to prove both of them is rather similar. Fitting's Theorem says the following,

Theorem:(Fitting's Theorem) Let $M$ and $N$ be normal nilpotent subgroups of a group $G$. If $c$ and $d$ are the nilpotent classes of $M$ and $N$, then $L=MN$ is nilpotent of class at most $c+d$.

In our case, these conditions all hold, we just have the additional conditions that $G=MN(=L)$ and that $M\cap N=1$, which makes our life easier. To prove Fitting's Theorem, start by proving that for $U, V, W\lhd G$ we have $[UV, W]=[U, W][V, W]$ and $[U, VW]=[U, V][U, W]$. The nilpotency of $L$ follows quickly. I will leave you to prove the class is $c+d$ on your own.

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Thanks Swlabr. I understand the computations in the first paragraph, (I take it $[(h_1,k_1),(h_2,k_2)]=([h_1,h_2],[k_1,k_2])$? But I don't understand how paragraph two follows. How do your computations come into play? Also, what is $G_i$? If $G=H\times K$, is $G_i=H_i\times K_i$? – yunone Jul 4 '11 at 14:22
Well, a group $G$ is nilpotent if and only if $G_i=\langle 1\rangle$ for some $i \in \mathbb{N}$. In the first paragraph I hinted towards showing, via induction, that $G_i \leq H_i \times K_i$, as (inductively) $G_{j}=[G, G_{j-1}]\leq [G, H_{j-1}\times K_{j-1}]= [H\times K, H_{j-1}\times K_{j-1}]\leq [H, H_{j-1}]\times [K, K_{j-1}]=H_j\times K_j$. As both $H$ and $K$ are nilpotent there exists some $i$ such that $H_i=\langle 1 \rangle=K_i$, and as $G_i\leq H_i\times K_1=\langle 1\rangle$ we are done... – user1729 Jul 4 '11 at 14:31
I see now, thanks much for the added explanation. – yunone Jul 4 '11 at 15:26
@Artus: I have thought about this a bit, and it is definately an equals sign. Plus, I have found a question in my notes from undergrad which (essentially) asks to show this with an equals sign. So definately $=$ not just $\leq$. – user1729 Jul 22 '13 at 9:19
Haven't you shown (the stronger result) that the direct product is nilpotent of class $\max c,d$? – Pete L. Clark Jul 22 '13 at 10:23

The class of all nilpotent groups of nilpotency class at most $c$ is exactly the collection of all groups for which the group word $$[x_1,x_2,x_3,\ldots,x_{c+1}]$$ evaluates to $1$ for all possible choices of $x_1,\ldots,x_{c+1}$ in the group (here, commutators are $[a,b]=a^{-1}b^{-1}ab$, and we write them left-normed, so that $[a,b,c] = [[a,b],c]$).

Since a collection of groups that is defined by group words forms a variety, and is closed under quotients, subgroups and arbitrary direct products, it follows that an arbitrary direct product of nilpotent groups of class at most $c$ is always nilpotent of class at most $c$.

If you have a finite collection $G_1,\ldots,G_n$ of nilpotent groups, then letting $c_i$ be the class of $G_i$, we can let $c=\max\{c_1,\ldots,c_n\}$; then each $G_i$ is nilpotent of class at most $c$, and thus the product is nilpotent of class at most $c$.

More generally any family of groups with bounded nilpotency class will have a product which is nilpotent (of class at most the common bound for the class).

On the other hand, taking a product of groups of unbounded nilpotency class will yield a group that is not nilpotent.

(The same idea works to show that any family of groups of bounded solvability class is solvable, and in particular that any finite family of solvable groups has solvable product).

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One nice characterization of finite nilpotent groups is that a finite group is nilpotent if and only if it is the (internal) direct product of its Sylow subgroups (or equivalently, every Sylow subgroup is normal). From this characterization the fact that you want follows almost immediately.

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Thanks Pete, so I could just then commute the factors in the direct product to get what I want. – yunone Jul 4 '11 at 9:04
@yunone: indeed. (Let me say though that I mostly just picked the characterization of finite nilpotent groups that I have always found easiest to remember. It is surely also possible to argue directly in terms of the central series stuff, and maybe is more educational to do so...but I haven't tried it myself.) – Pete L. Clark Jul 4 '11 at 9:23
My point being that another, more direct, answer to your question would be very welcome. – Pete L. Clark Jul 4 '11 at 9:29
Yes, I'd be glad to see one like that too. But thanks for this other characterization, and +1 in the meantime. – yunone Jul 4 '11 at 9:36

Here's a proof which works for finite or infinite nilpotent groups. If $G = G_1 \times G_2 \times \ldots \times G_n$, then $Z(G) = Z(G_1) \times \ldots \times Z(G_n)$, so that $G/Z(G) \cong G_1/Z(G_1) \times \ldots \times G_n/Z(G_n)$. Continuing in this way, you can see that if the upper central series for each $G_i$ eventually reaches $G_i$, (ie, if each $G_i$ is nilpotent), then the upper central series for $G$ eventually reaches $G$ (ie, $G$ is nilpotent).

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