TITLE: Set of left cosets and set of right cosets of a not normal subgroup could be equal?
QUESTION [5 upvotes]: Question 1:
Let $H$ be a subgroup of a finite group $G$. I would like to show an example for H, which is not normal, but the partition of $G$ into the left cosets of H is equal to the partition of $G$ into the right cosets of H.
Example:
A definition of the Dihedral group of order 6 is $D_3=gp\{c,b\}\ ,\ c^3=b^2=(bc)^2=e$. It follows from the definition, that $D_3=\{e,c,c^2,b,bc,bc^2\}$, and $bc=c^2b\ ,\ cb=bc^2$. 
$H=\{e,b\}$ a subset of $D_3$.
H is not normal because $\exists c \in D_3: \  cH=\{c,bc^2\} \neq \{c,bc\}=Hc$
But the set of left cosets and the set of right cosets is the same:
Set of left cosets: $L=\{gH\ |\  \forall g \in D_3\}=\Big\{\{e,b\},\{c,bc^2\},\{c^2,bc\}\Big\}$ 
Set of right cosets: $R=\{Hg\ |\  \forall g \in D_3\}=\Big\{\{e,b\},\{c,bc\},\{c^2,bc^2\}\Big\}$  
So $R\neq L$
Refer to "coffeemath" 's answer a missed the calculation at first time, now i correct it. So it's not a counter example for Question 2.
Question 2:
Is the next statement is true? 
H is not normal $\ \Rightarrow \ \exists a \in G : aH\neq Hb \quad \forall b \in G $
I only see, that (from the negate of the definition of a normal subgroup)
H is not normal $\ \Rightarrow \ \exists a \in G : aH\neq Ha $

REPLY [4 votes]: If $aH=Hb$ then $a=ae\in aH=Hb$ so that $Hb=Ha$. 
So if the partitions of left- and of right-cosets coincide then $aH=Ha$ for every $a$ and consequently $H$ is normal.