Step in Proof of Cardinality of Product of two Groups

Let $A,B < G$ be two subgroups of some group $G$, then I have a question on the proof of the following:
$$
|AB| = \frac{|A||B|}{|A \cap B|}.
$$
Proof: Let $D = A \cap B$, arrange $A$ and $B$ in left cosets and right cosets regarding $D$ (which is also a subgroup)
$$
A = D \cup s_2 D \cup s_3 D \cup \ldots \cup s_n D, \quad
B = D \cup D t_2 \cup D t_3 \cup \ldots \cup D t_m
$$
Now the fact that $\frac{|A||B|}{|D|}$ are distinct results from the following equations:
If
$$
s_{\alpha}D D t_{\beta} = s_{\alpha_1} D D t_{\beta_1}
$$
then
$$
s_{\alpha_1}^{-1} s_{\alpha} D = D t_{\beta} t_{\beta}^{-1}
$$
As the first member of the last equation represents an element of $A$ while the second member represents an element of $B$, it results that the last equation implies $\alpha_1 = \alpha$ and $\beta_1 = \beta$.

The last step is not clear to me, why does $\alpha_1 = \alpha$ and $\beta_1 = \beta$ follows? Is it the case that if $aD = Db$ for $a \in A$ and $b \in B$ that $a=b=1$ follows? I don’t believe this and cannot prove it either? Could someone please explain the last step?

Solutions Collecting From Web of "Step in Proof of Cardinality of Product of two Groups"

It is not true in general that $aD = Db$, $a \in A$, $b \in B$, implies $a = b$. In fact $aD = Db$ is true for any choice of elements $a, b \in D$. What’s being used here is the fact that the $s_\alpha$ and $t_\beta$ are a set of representatives for the cosets. This means if $s_\alpha \neq s_{\alpha_1}$ then $s_\alpha D \neq s_{\alpha_1}D$. Turning this around, if $s_\alpha D = s_{\alpha_1}D$ then we must have $s_\alpha = s_{\alpha_1}$.

Now, you know that $s_{\alpha_1}, s_\alpha \in A$ and $D \subseteq A$ so $s_{\alpha_1}^{-1}s_\alpha D \subseteq A$. Similarly, $Dt_\beta t_{\beta_1}^{-1} \subseteq B$. As these are equal you get $s_{\alpha_1}^{-1}s_\alpha D \subseteq A \cap B = D$ and hence $s_{\alpha_1}^{-1}s_\alpha D = D$. But this can be rewritten as $s_{\alpha_1}D = s_\alpha D$, therefore $s_{\alpha_1} = s_\alpha$.

A similar argument works for the $t_\beta$.