# About the definition of powerful p-groups

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.

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.
• 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 9 '19 at 11:52