Prove that any two left cosets $aH, bH$ either coincide or are disjoint, and prove Lagrange's theorem

Solution 1:

Since none has answered so far, I will post my comments as an answer.

$aH$ is the set $\{ah: h\in H\}$ therefore once you found that $a=bh_2h_1^{−1}$, then for every $h\in H$ you have $ah=bh_2h_1^{−1}h\in bH$ (since $h_2h_1^{−1}h$ still lies in $H$). Therefore $aH\subseteq bH$; the same reasoning shows that the other inclusion holds too, so you get the implication $$ aH\cap bH\neq\emptyset\quad\Leftrightarrow\quad aH=bH $$ (the '$\Leftarrow$' follows from the fact that aH and bH are not empty; does it answer your first question?).

For every $g\in G$ you have $g\in gH$ so $G=\bigcup_{g\in G}gH$. As proved above, for every $g_1,g_2\in G$ $g_1H$ and $g_2H$ either are disjoint or they coincide, therefore there exist $g_1\ldots g_n\in G$ such that $$ G = \bigcup_{i=1}^n g_iH \qquad\text{and}\qquad g_1H\ldots g_nH \text{ are pairwise disjoint} $$ Each set $g_iH$ has the same cardinality as $H$ (trivially for every $h_1,h_2\in H$ you have $h_1\neq h_2$ iff $g_1h_1\neq g_1h_2$), so the above formula implies $$ |G| = \sum_{i=1}^n|g_iH| = \sum_{i=1}^n|H| = n|H| $$ (This should explain your last question)

Therefore the number $n$ of 'representatives' does not depend on the elements chosen and is by definition $n=|G:H|$. So $|G|=n|H|=|G:H||H|$.