Exercise 1.2.11 in Tao's "Hilbert's Fifth Problem"

Question

I am reading Tao's book "Hilbert's fifth problem" (freely available online https://terrytao.files.wordpress.com/2012/03/hilbert-book.pdf) and I'm trying to do most exercises. I have little experience with calculations involving asymptotics, hence I want to ask if the following solution is correct. The exercise I'm trying to solve is 1.2.11 (ii): Let $V$ be a $C^{1,1}$ local group. Show that $$x*y*x^{*-1}*y^{*-1} = O(|x||y|),$$ for $x,y$ sufficiently close to the identity.

I recall the definition of a $C^{1,1}$ local group.

Definition: A $C^{1,1}$ local group is a local group $V$ which is a neighbourhood of the identity $0$ in a Euclidean space with group identity $0$, whose operation $*$ obeys $$x*y = x+y +O(|x||y|),$$ for $x,y$ sufficiently close to the identity.

My solution

We follow the hint and start by proving $x*y = y* x + O(|x||y|).$ Indeed, \begin{align*} |x*y| &\leq |x+y| + C|x||y| \\ &\leq |y*x| + C|x||y| + C|x||y|\\ &\leq |y*x| + C|x||y|. \end{align*} Now we calculate \begin{align*} x*y &= (x*y*x^{*-1}*y^{*-1}) * (y *x) \\ &\stackrel{?}{=}(x*y*x^{*-1}*y^{*-1}) * (x *y) + O(|x||y|) \\ &= x*y*x^{*-1}*y^{*-1} + x *y + O(|x||y|) + O(|x*y*x^{*-1}*y^{*-1}||x*y|). \end{align*} Thus \begin{align*} 0 &= x*y*x^{*-1}*y^{*-1} + O(|x||y|) + O(|x*y*x^{*-1}*y^{*-1}||x*y|) \\ &= x*y*x^{*-1}*y^{*-1} + O(|x||y|), \end{align*} for $x,y$ sufficiently close to the origin.

Question

Is this solution valid? I'm particularly worried about the equality indicated with a question mark.

Answer 1

Following Tao's hint, let $z = x*y*x^{*-1}*y^{*-1}$. Then, for $x, y$ near $0$, $$x*y = z * y * x = z + y*x + O(|z|\,|y*x|).$$ Both $x*y$ and $y*x$ are $x+y+O(|x|\,|y|)$, and $|y*x| = O(|x|+|y|)$, so it follows that $$z = O(|x|\,|y| + |x|\,|z| + |y|\,|z|).$$ In other words there are constants $\delta>0$ and $C$ such that whenever $|x|,|y| < \delta$ we have $$|z| \le C(|x|\,|y| + |x|\,|z| + |y|\,|z|).$$ Let $\delta' = \min(\delta, 1/(4C))$. Then for $|x|, |y| < \delta' \le 1/(4C)$ we have $C (|x|\,|z|+|y|\,|z|) \le |z|/2$, so the inequality above implies $|z| \le 2C|x|\,|y|$, i.e., $|z| = O(|x|\,|y|)$.