What would be an example such that $aH=bH$ but $Ha \neq Hb$? 
Let $H$ be a subgroup of a group $G$.
Assume $aH=bH$ for some $a,b\in G$
What would be an example such that $Hb\neq Ha$?

I cannot imagine what would be.. Moreover, if $H$ is infinite, is this still not true?
 A: Note $aH=bH\Leftrightarrow b^{-1}a\in H$ and $Ha\ne Hb\Leftrightarrow ab^{-1}\not\in H$. This signifies we might be able to find an example by picking $a$ and $b^{-1}$ that do not commute, and wlog making $H=\langle b^{-1}a\rangle$, but we still need to make sure $ab^{-1}$ is not a power of $b^{-1}a$, which is the case if $a$ and $b$ satisfy no relations.
Let $G$ be the free group on $\{a,b\}$ and let $H=\langle b^{-1}a\rangle$ be the cyclic group generated by $b^{-1}a$.
Then $b^{-1}a\in H\implies aH=bH$ however $Ha=Hb\implies ab^{-1}\in H$ which can't be the case because every element of $H$ is a word in the letters $a$ and $b$ beginning with either $b^{-1}$ or $a^{-1}$.
A: Let us consider this finite group of order $8$, call it $G$. Then $H=\{0,1\}$ is a subgroup of $G$, but $H$ is not normal. And, there, we have:
$$3H=\{3,6\}=6H,$$
but
$$H3=\{2,3\}\ne\{4,6\}=H6.$$
Similarly, in the same environment, we have:
$$H2=\{2,3\}=H3,$$
but
$$2H=\{2,4\}\ne\{3,6\}=3H.$$
A: (This is MattAllegro's answer rewritten in the context of $D_4$, the symmetry group of a square.)

Working in $D_4$, let $r$ and $f$ be a quarter rotation and a reflection.  Then $H= \{ e, f\}$ is a subgroup of $D_4$.  We have
$$rH = \{ r, rf \} = \{ rf, r \} = rfH $$
but
$$ Hr = \{ r , fr \} \neq \{fr,frf\} = Hrf .$$
