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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!)

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In what sense is the wreath product not elementary? –  Qiaochu Yuan Dec 14 '10 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 '10 at 4:53
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@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 '10 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 '10 at 5:06
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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 '10 at 11:56

1 Answer 1

up vote 14 down vote accepted

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.)

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Nice! I thought of this example but had to eat dinner before I could prove it worked. –  Qiaochu Yuan Dec 14 '10 at 5:12
    
Brilliant! I was hoping some sort of matrix fiddling like this would do the trick. I guess I'm not sure how much more elementary this is than the example I gave, but at least the group can be described in more elementary language. This example is also cleverer, I think. Thanks! –  Zach N Dec 14 '10 at 5:21
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For another proof that the center is trivial, see Remark 5.10 in math.uconn.edu/~kconrad/blurbs/grouptheory/conjclass.pdf –  KConrad Dec 14 '10 at 11:51

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