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I am reading "Analytic pro-$p$ groups" by Dixon, Du Sautoy, Mann and Segal.

They define $G$ a finite $p$-group to be powerful if $[G,G]\leq G^p$ for $p$ odd but in the case $p=2$ they require $[G,G]\leq G^4$.

Why this discrepancy between odd and even $p$? Is it related to some substantial fact about $p$-groups or is this just a technical condition you assume to make the proofs work?

This reminds me of the computation $\mathbb{Z}_p^{\times}=\mathbb{F}_p^{\times}\times \mathbb{Z}_p$ for $p$ odd while for $p=2$ we have $\mathbb{Z}_2^{\times}=\{ \pm1\}\times \mathbb{Z}_2$. But I could not correlate directly these two facts.

Shaun
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N.B.
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1 Answers1

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The condition $[G,G]\leq G^2$ is vacuous: it holds for any group. Because $G/G^2$ is a group of exponent $2$, and hence abelian, so $G^2$ contains $[G,G]$.

The point of the powerful condition is that it gives you control over expressions of $p$th powers, akin to the case of regular $p$-groups (though it turns out to be a much more useful property than regularity). Since $[G,G]\leq G^2$ holds trivially, the concept does not give you enough control over the group. Requiring $[G,G]\leq G^4$ plays the analogous role for $2$-groups, giving you enough to work on.

Arturo Magidin
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  • I thought that the core of the problem was about some property of the p-groups which I was ignoring. I really did not consider the basic fact that a group of exponent $2$ is abelian. Thanks for the answer. – N.B. Nov 09 '19 at 11:52