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

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 $aH \cap bH = \emptyset$ (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 the theorem?

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I believe that $aH = \{ah_1\} = \{b(h_2h_1^{-1})h_1\} \in H$ should read as $aH = \{ah_1\} = \{b(h_2h_1^{-1})h_1\} \in bH$. That should help clarify your first part. –  Calvin Lin Dec 29 '12 at 16:45
Mistyped $aH \cap bH = \emptyset$ for $aH \cap bH \neq \emptyset$... –  AndreasT Dec 29 '12 at 16:49
It was correct initially. They are not disjoint, so their intersection is NOT the empty set which has notation $\emptyset$ –  Calvin Lin Dec 29 '12 at 16:54
$aH$ 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, so you get the implication $aH\cap bH\neq \emptyset \Leftrightarrow aH=bH$ (the '$\Leftarrow$' follows from the fact that $aH$ and $bH$ are not empty!). –  AndreasT Dec 29 '12 at 16:59
For every $g\in G$ you have $g\in gH$ so $g=\bigcup_{g\in G} gH$. As proved above, $g_1H$ and $g_2H$ either are disjoint or they coincide, therefore there exist $g_1\ldots g_n\in G$ such that $$G=\bigcup_{i=1}^n g_iH \quad\text{and g_1H\ldots g_nH are pairwise disjoint}$$ Each set $g_iH$ has the same cardinality as $H$ (trivially for every $h_1,h_2\in H$ $h_1\neq h_2$ iff $g_1h_1\neq g_1h_2$), so the above formula implies $$|G|=\sum_{i=1}^n |g_iH| = \sum_{i=1}^n |H| = n|H|$$ Therefore the number $n$ of 'representatives' does not depend on the elements chosen and is by definition $|G:H|$. –  AndreasT Dec 29 '12 at 17:12
$aH$ 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 $|G|=n|H|=|G:H||H|$.