# 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 $t_\beta$.