Let's pick a specific group, so we can be definite about this. To make things easier, we'll pick a rather small group.
Specifically, let:
$G = \{e,a,a^2,b,ab,a^2b\}$ where:
$a^3 = b^2 = e$, and the multiplication is given by the rule:
$ba = a^2b$.
Now we need to pick a subgroup, so let $H$ be the subgroup:
$H = \{e,b\}$.
Let's look at the left cosets first:
$eH = H = \{e,b\}$.
$aH = \{a,ab\}$
$a^2H = \{a^2,a^2b\}$
$bH = \{b,e\} = H$
$abH = \{ab,a\} = aH$
$a^2bH = \{a^2b,a^2\} = a^2H$
so we have 3 different (distinct) left cosets: $H,aH,a^2H$.
Now, let's look at the right cosets:
$He = H = \{e,b\}$
$Ha = \{a,ba\} = \{a,a^2b\}$ (see above)
$Ha^2 = \{a^2,ba^2\} = \{a^2,ab\}$
(because $ba^2 = (ba)a = (a^2b)a = a^2(ba) = a^2(a^2b) = a^4b = (a^3)(ab) = ab$).
$Hb = \{b,e\} = H$
$Hab = \{ab,bab\} = \{ab,a^2\} = Ha^2$
$Ha^2b = \{a^2b,ba^2b\} = \{a^2b,a\} = Ha$
Again, we have three distinct right cosets, as well: $H,Ha,Ha^2$.
But notice these are DIFFERENT sets than the left cosets, for while:
$eH = H = He$
$bH = H = Hb$
we have:
$aH = \{a,ab\}$ which has $ab$ as a member, whereas:
$Ha = \{a,a^2b\}$, which does not have $ab$ as a member.
Now this is bad news if we were planning on trying to define:
$(Ha)(Hb) = Hab$.
Let's look at what the set $(Ha)(Hb) = \{xy: x\in Ha,y \in Hb\}$ actually works out to be. We have 4 possible products:
$ae,ab,(a^2b)(e),(a^2b)(b)$, and working out what these are gives us:
$ae = a$
$ab = ab$
$a^2b = a^2b$
$(a^2b)(b) = a^2$
so $(Ha)(Hb) = \{a,a^2,ab,a^2b\}$ and this isn't even ANY coset of $H$ (all the cosets of $H$ have 2 elements, the same number of elements $H$ has, but this product set has FOUR elements).
