Skip to main content
Post Closed as "Needs details or clarity" by YCor, Lucia, Jeremy Rickard, Marco Golla, Alex Degtyarev
added 436 characters in body
Source Link
john mangual
  • 22.8k
  • 4
  • 63
  • 172

I am trying to understand the concept of approximate group.


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 $o(e)$?


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.

I am trying to understand the concept of approximate group.


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 $o(e)$?

I am trying to understand the concept of approximate group.


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 $o(e)$?


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.

Source Link
john mangual
  • 22.8k
  • 4
  • 63
  • 172

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

I am trying to understand the concept of approximate group.


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 $o(e)$?