Welcome to MathOverflow

Question

9

1

The only examples I have encountered of infinite $p$-groups with trivial center employ non-elementary methods in their construction. For instance, Example 9.2.5 of Scott's Group Theory is a perfectly satisfactory example, but it requires the wreath product (which, though an invaluable group-theoretic tool, is not what I consider an "elementary method").

Does anyone know of an example (of an infinite $p$-group with trivial center) that can be constructed and proven to have the claimed properties in a way that is friendly to, say, students of a first course in group theory? Perhaps a large product of finite groups or an easy-to-describe matrix group?

(I also welcome arguments for the nonexistence of such an example!)

flag	asked Dec 14 2010 at 4:27 Zach N 196●7

1

In what sense is the wreath product not elementary? – Qiaochu Yuan Dec 14 2010 at 4:32

@Qiaochu: I just meant that a typical first course in groups doesn't cover the wreath product, though it would be useful for students in such a course to know (and perhaps see a proof) that the statement Every $p$-group has nontrivial center is false for infinite $p$-groups. Other natural generalizations in basic group theory have accessible counterexamples. The one that comes to mind is A group in which every nontrivial element has order 2 is abelian. Changing '2' to '3' gives a false statement, and there is a not-too-hard-to-describe group of order 27 that illustrates this. – Zach N Dec 14 2010 at 4:53

1

@Zach: If the course includes semi-direct products, then wreath products are an important special case. If it includes Sylow subgroups, then there's a lot to be said for showing them what the $p$-Sylow subgroup of the $p^n$th symmetric group is. That's another, maybe even better, reason for mentioning wreath products. – Tom Goodwillie Dec 14 2010 at 5:00

@Tom: Excellent point! And apologies to Qiaochu since my example wasn't fair; I concealed "hard-enough-to-require-semidirect-products" in "not-too-hard-to-describe". – Zach N Dec 14 2010 at 5:06

2

Zach, concerning your comment about changing 2 to 3, I think it'd be better to change 2 to any odd prime p: in the (nonabelian) Heisenberg group over Z/p for odd p, each non-identity element has order p. – KConrad Dec 14 2010 at 11:56

show 1 more comment

Emerton · Accepted Answer · 2010-12-14 04:59:26Z UTC

Here is a matrix example, hopefully correct and also hopefully sufficiently simple:

Consider the group $U$ of $\infty\times \infty$ upper unipotent matrices with entries in $\mathbb F_p$, with all but finitely many entries equal to zero; so they have the form $$\begin{pmatrix} 1 & a_{1 2} & a_{1 3} & \cdots \\ 0 & 1 & a_{2 3} & \cdots \\ \vdots & \vdots & \vdots \end{pmatrix},$$ with $a_{i,j} \in \mathbb F_p$ and all but finitely many $a_{i j } = 0$.

Note that $U$ admits a homomorphism onto the upper unipotent matrices $U_n$ in $GL_n(\mathbb F_p)$ for any $n$, given by forgetting the $a_{i,j}$ for $i$ or $j$ greater than $n$. If $z$ is in the centre of $U$, then its image lies in the centre of $U_n$ for each $n$. But the centre of $U_n$ has trivial image in $U_{n-1}$, and so the centre of $U$ actually has trivial image in each $U_n$. Since $U$ evidently embeds into the projective limit of the $U_n$, we see that $U$ has trivial centre. (Intuitively, we've made the nilpotency class infinite, and so pushed the centre away to infinity.)

Simple(st) example of an infinite $p$-group with trivial center

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.