<s>[INST] Left and Right Cosets Let $H$ be a subgroup of a group $G$ and let $a, b \in G$. I need to give a counterexample or proof of the following statement:
If $aH = bH$, then $Ha = Hb$
Proof: 
For every $h \in H, ah = bh$
$
\begin{align}
ah &= bh \newline
ahh^{-1} &= bhh^{-1} \newline
a &= b \newline
ha &= hb
\end{align}
$
Could someone critic my proof?
Thanks in advance.

Edit
Look at the answer below.
 [/INST] Counterexample: We want to use a non-Abelian group such as $S_{3}$
$\mu_{1} = \pmatrix{1&2&3\\1 &3 &2}$, $\mu_{2} = \pmatrix{1&2&3\\3 &2 &1}$, 
$\mu_{3} = \pmatrix{1&2&3\\2 &1 &3}$
$\rho_{0} = \pmatrix{1&2&3\\1 &2 &3}$, $\rho_{1} = \pmatrix{1&2&3\\2 &3 &1}$, 
$\rho_{2} = \pmatrix{1&2&3\\3 &1 &2}$
Let $H = \{\rho_{0}, \mu_{2}\}$, $a = \mu_{1}$ and $b = \rho_{1}$
$aH = \mu_{1}\{\rho_{0}, \mu_{2}\} = \{\mu_{1}, \rho_{1}\}$
$bH = \rho_{1} \{\rho_{0}, \mu_{2}\} = \{\rho_{1}, \mu_{1}\}$
But 
$Ha = \{\rho_{0}, \mu_{2}\}\mu_{1} = \{\mu_{1}, \rho_{2}\}$
$Hb = \{\rho_{0}, \mu_{2}\}\rho_{1} = \{\rho_{1}, \mu_{3}\}$
So $Ha \neq Hb$
</s>