Prove Equivalence Relation in G Hei, guys!
I'm having some trouble with the next problem:
Let $A$ and $B$ be subgroups of $G$. Show that $\sim$ is an equivalence relation when it is
defined as follows:
$g\sim g'\Leftrightarrow g' = agb$ for some $a$ from $A$, $b$ from $B$.
Should I prove this is valid for any a and b? It says for some elements a and b. But I might as well take the unit element. 
Could you give me a clue on how to solve it?
Cheers!
 A: We can show that by verifying 3 properties:


*

*Reflexive- $x\sim x$ by using $a = b =e$ which obviously exists since $A,B$ are subgroups and therefore, $e \in A,B$.

*Symmetric - $x\sim y$ then there exist $a\in A$ and $b\in B$ s.t. $y = axb$. $A, B$ are subgroups and therefore $a^{-1}\in A$ and $b^{-1}\in B$ and therefore $x = a^{-1}yb^{-1}$ which imply $ y\sim x$

*Trasitive - $x\sim y$ and $y\sim z$ then there are $a,a'\in A, b,b'\in B$ s.t. $y=axb$ and $z=a'yb'$ and by subtitution of $y$ we get $z = aa'xb'b$ and because $A,B$ are subgroup and closed under multiplextion we get that $aa'\in A$ and $bb'\in B$ and therefore $x\sim z$

A: The problem says $ g \sim g' $ if and only if there are elements $ a \in A $ and $b \in B$ such that $ g' = agb  $
Reflexivity: For this the unit element which is present in both $A$ and $B$ is exactly what you need. 
Symmetry: We need to show that if $ g \sim g'$ then $g' \sim g$. That is if there are elements $a \in A, b \in B $ such that $ g' = agb $ then there are elements $ a' \in A, b' \in B $ such that $ g = a'g'b' $. Try using the inverse elements of $a$ and $b$
Transitivity:  We are given that $ g \sim g' $ and $g' \sim g'' $ from which we need to prove that $ g \sim g'' $. From the assumption there are elements $ a, a' \in A $ and $b, b' \in B$ such that $g' = agb $ and $ g'' = a'g'b' $. Think of the most obvious substitution. Then think why there must be elements $ a'a \in A $ and $b'b \in B$ which give us the required relation between $g$ and $g''$
A: i. Reflexivity 
$g \sim g$.  $\ \ \ $Indeed, for $\ \exists\ \  e_{A} \in A\ \    \text{and}\ \  e_{B} \in B\ \  \text{such that}\ \  g= e_{A}\ g\ e_{B} \Rightarrow g = g$. 
ii. Symmetry 
$g \sim g' \Rightarrow g'\sim g$. $\ \ \ $ Since there are $a \in A $ and $b \in B$ such that $g = a\ g\ b$ then as $A$ and $B$ are subgroups, there exist $a^{-1} \in A$ and $b^{-1} \in B$ such that $g=a\ g'\ b \Rightarrow g' = a^{-1}\ g \ b^{-1}   \Rightarrow g' \sim g $. 
iii. Transitivity 
$g \sim g''$ and $g'' \sim g' \Rightarrow g \sim g'$. $\ \ \ $In fact, there exist $a',a'' \in A$ and $b',b'' \in B$  such that $g = a''\ g''\ b''$ and $g'' = a'\  g'\ b'$, then $g = a'' \ (a'\ g' \ b') \ b'' \Rightarrow g = (a' a'')\ g'\ (b' \ b'')$. Let $a : = a'\ a'' \in A$, and $b := b'\ b'' \in B$ then we have $g = a\ g'\ b$ which means that $g \sim g'$. 
A very important notion in Groups lies on this equivalence relation as follow 
$$ y \sim x \Leftrightarrow \exists h \in H\ \  \text{such that}\ \ y = xh$$
As an exercise you could show that this is an equivalence relation. This set $ \lbrace y \in G ; y \sim x \rbrace = \lbrace xh ; h \in H \rbrace$ is denoted by $xH$ and is called left coset. 
