Consider a group $G$ and a subgroup $H \subset G$. Prove that any two left cosets $aH$ and $bH$ either coincide or are disjoint. State and prove the Lagrange theorem.
For the first part, the proof I have goes like this:
Assume $aH \cap bH \ne \emptyset$ (so are not disjoint). Then for some elements $h_1, h_2 \in H$, we get
$ah_1 = bh_2.$
Multiply both sides on the right by $h_1^{-1}$
$ah_1h_1^{-1} = bh_2h_1^{-1}$ $a = b(h_2h_1^{-1})$
I get up to here, but then for some reason the next line says
$aH = \{ah_1\} = \{b(h_2h_1^{-1})h_1\} \in H$
As all elements of $H$ here, this is equal to $aH$ and so the proof is complete.
I don't get why the intersection giving the empty set shows its not disjoint.
Secondly, to prove the Lagrange theorem. The theorem is that the order of the subgroup divides the order of the group, or:
$|G| = |H| \cdot (\mathrm{Number \, of \, left \, cosets\, of H)}$
Then the proof says this:
$G$ consists of the number of left cosets of $H$, and each of them consist of $|H|$ elements, then the cosets are disjoint.
How does this prove Lagrange's theorem?