An approximate version of $g^2 = e$ for all $g \in G$, implies $G$ is Abelian [closed]

Question

I am trying to understand the concept of approximate group.

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So I took a group theory exercise from a physics class at Caltech. The question basically states:

Suppose that for any element $g \in G$ we have $g^2 = e$, then $G$ is Abelian, i.e. $g_1 g_2 = g_2 g_1$ for all $g_1, g_2 \in G$.

In other words, every element is a reflection or the identity implies, $G = (\mathbb{Z}/2\mathbb{Z})^n$.

Let's try to write down an approximate version of this exercise using little-o notation.

Suppose that for any group element $g \in G$ we have $g^2 = o(e)$, then $G$ is nearly Abelian, i.e. ???

Now let's follow the proof. We only have that $g^{-1} \approx g$

• $(g_1 g_2)^2 = g_1 g_2 g_1 g_2 = o(e)$
• $ g_2 g_1 = o(g_1^{-1} e g_2^{-1}) \approx o(g_1 e g_2) = g_1 g_2 \, o(e)$

This doesn't look very Abelian, and I sort of made things up as far as how $o(e)$ should behave:

$$ o(e) = \text{neighborhood of the identity}$$

and hopefully $o(e)^2 \approx o(e)$.

Sorry if this is too open-ended or unclear. In real math we deal with things which are not-quite symmetries and not-quite groups. How do we formalize such a situation, regarding $o(e)$?

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The Ben Green article gives us 3 options for defining an approximate group, $A$ each of them related to ideas from abstract algebra class:

• $\mathbb{P}[xy^{-1} \in A] > \frac{1}{K} $
• $|A^2| \leq K|A|$
• $A^2$ can be covered by $K$-right translates of $A$.

The first definition sort of makes sense to me. Another possibility is that I have invented my own definition of approximate group, deserving its own term.

Answer 1

As suggested by Igor Rivin's argument, it is the case that if we put the operator norm (with respect to Euclidean norm on vectors) on $M_{n}(\mathbb{C})$ and consider a finite subgroup $G$ of $U_{n}(\mathbb{C})$ such that $\|g^{2}-e \| < 1$ for all $g \in G$, then $G$ is Abelian.

Later edit: I realise that there is a much simpler argument than my original one, with a sharper bound. Note that $M$ is a matrix of finite order in ${\rm GL}(n,\mathbb{C})$ with $\| I - M^{j} \| < 1$ for all $j$, then for any eigenvalue $\lambda$ of $M$, we have $|1-\lambda^{j}| < 1$ for all $j$, so that $\lambda^{j}$ has strictly positive real part for all $j$. But $1$ is the only complex root of unity with the property that all of its powers have positive real part ( since the sum of all powers of any other root of unity is $0$). Hence $M = I$.

Hence if $\|I -g^{2} \| < 1$ for all $g \in G$, where $G$ is a finite subgroup of ${\rm U}_{n}(\mathbb{C})$, then $g^{2} = I$ for all $g \in G$, so that $G$ is Abelian.

The second part of the answer is somewhat tangential to the original question, and follows the direction suggested by Sean Eberhard's comment.

For a finite group $G$, it is the case that if more than $\sqrt{\frac{5}{8}} |G|$ elements $x \in G$ have $x^{2} = e$, then $G$ is Abelian. The dihedral group of order $8$ ( I mean the one with $8$ elements) - and direct products of it with elementary Abelian $2$-groups as large as you like-show that this can't be improved much as a general bound, since a dihedral group $D$ of order $8$ contains $6$ elements which square to the identity and $ 6 < \sqrt{\frac{5}{8}} |D| <7$ in that case.

This is because (as noted in the paper linked to in Sean Eberhard's comment, and also previously noted by Brauer and Fowler), the count of solutions to $x^{2} = e$ given using the Frobenius-Schur indicator leads easily to $\sqrt{\frac{5}{8}} |G| < \sqrt{k(G)}\sqrt{|G|}$ in the case under consideration, where $k(G)$ is the number of conjugacy classes of $G$. Hence $\frac{k(G)}{|G|} > \frac{5}{8}$, so the probability that two elements of $G$ commute is greater than $\frac{5}{8}$, in which case $G$ is Abelian by a Theorem of W.Gustafson. If preferred, this can be seen directly using character theory- in general, if $G$ has $k$ conjugacy classes, we have $[G:G^{\prime}] + 4(k - [G:G^{\prime}]) \leq |G|$ by the orthogonality relations. Hence $\frac{k}{|G|} \leq \frac{1}{4} + \frac{3}{4|G^{\prime}|} \leq \frac{5}{8}$ if $G^{\prime} \neq 1$, ie if $G$ is non-Abelian.

Answer 2

The commutator of two elements is the product of three squares (exercise) [that gives a scientific proof of the initial theorem]. So, what you are saying is that the commutator subgroup is contained in a small neighborhood of the identity, which, for Lie groups (which are more or less the same as matrix groups), would imply that the commutator subgroup is trivial. If you relax the meaning of "small", then you get nilpotent groups, and such.