$G$ group, $A,B$ sub groups. $R$ binary relation, $xRy \iff \exists a\in A, b\in B$ such that $ y = axb $

Question

$G$ is a group with sub-groups $A,B$ in it. Lets define a binary relation $R$ in $G$ in that way: $xRy \iff \exists a\in A, b\in B$ such that $ y = axb $.

First thing that I'm not sure how to approach: Prove $R$ is reflexive, symmetric, trans.

Well, I don't understand how the binary relation works, so I've written so far

$(1)$ $xRx \rightarrow (x=axb) \in G$, thus it is reflexive.

$(2)$ $xRy \rightarrow (y=axb) = (x = ayb) \rightarrow yRx$, thus it is symmetric.

$(3)$ $x,y,z \in R$, $xRy \rightarrow y = axb$, $yRz \rightarrow z = ayb$. Thus,

$xRz \rightarrow z = a(axb)b = axb$ (not sure about this one)

This is purely by definition, so I hope that works.

Second thing that I'm not sure how to approach: Another thing is, how can I find the $[x]_R$, when $x \in G$ and $AxB = \{axb | a \in A, b \in B\}$, note that $AxB$ is a double coset, which I guess can say something about the index of the $[x]_R$ I guess index $2$ and thus I need to find only two elements in $[x]_R$? Not sure about that either.. would love some assistance.

Answer 1

Your "proofs" are not proofs for two reasons:

1. They're backwards: in "1", for instance, you start with $xRx$ and derive something from it, which is called "assuming what you're trying to prove", and is definitely not a valid method of proof.

2. Continuing with $1$, saying that $xRx$ doesn't say that $x = axb$. It says that there exists $a \in A$ and $b \in B$ with this property. Without those quantifiers, the statement that $x = axb$ means nothing, since $a$ and $b$ don't yet have a meaning.

Here's a revised proof for the first one. Notice how many words I write, and how each claim is justified with supporting argument:

We want to show that for any $x \in G$, we have $x R x$. Let $x$ be any element of $G$. Let $a = e$ and $b = e$, and note that $a \in A$ (because every subgroup contains the identity element) and $b \in B$. Further note that $x = e x e = a x b$, so we've shown the existence of elements $a,b$ of $A, B$ respectively such that $x = a x b$, hence we've show that $xRx$.

Here's a start for the second:

Suppose that $x, y \in G$ and $x R y$. Then by definition, there are elements $a \in A$ and $b \in B$ with $$ y = a x b $$ We want to show that $y R x$, i.e., we want to exhibit elements $a' \in A, b' \in B$ such that $$ x = a' y b'. $$

Let $a' = $ (something) and let $b' = $ (something). Then $a' y b' = $ (simplifications here) $ = x$, and we are done.