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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?

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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.2012-12-29
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    Mistyped $aH \cap bH = \emptyset$ for $aH \cap bH \neq \emptyset$...2012-12-29
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    It was correct initially. They are not disjoint, so their intersection is NOT the empty set which has notation $\emptyset$2012-12-29
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    $aH$ 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!).2012-12-29
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    @AndreasT Oh yeah, in my notes is says not equal. Ok, I understand that bit now about them being not empty, but how does this prove the Lagrange theorem? Or have I got two bits of the proof mixed up?2012-12-29
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    @CalvinLin Two sets $A$ and $B$ are disjoint iff their intersection IS the empty set, otherwise they share at least an element... It is the plain definition, isn't it?2012-12-29
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    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 $|G:H|$.2012-12-29
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    @AndreasT: You should put that as an answer :)2012-12-29
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    @AndreasT Currently the way it is written, the intersection of aH and bH is the empty set, so it means they are disjoint. However, the brackets after it says that (so are not disjoint). The following statement that $ah_1 = bh_2$ gives an element of $aH$ that is equal to $bH$. Hence, something is wrong with the statements as is, and should be corrected.2012-12-29
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    @CalvinLin I was remarking that either one of '$aH\cup bH=\emptyset$' or '(so are not disjoint)' is incorrect. Probably you did too, and if so I'm sorry I misinterpreted your comment :)2012-12-29
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    @Clayton Actually my answer lacks the first request, since I just give my proof and don't explain his...2012-12-29

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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 $|G|=n|H|=|G:H||H|$.

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    I think $g=\bigcup_{g\in G}gH$ should be $G=\bigcup_{g\in G}gH$, no?2016-03-11
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    Thanks @ImATurtle, I fixed it.2016-03-11
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    doesnt $ah=bh_2h_1^{−1}h$ automatically means that $aH=bH$ if we know that any coset $hH : h \in H = H$?. since $h_2h_1^{−1} = h_3$ and $h_3 h : h \in H$ is the same as $hH$ which is the same as $H$ since $h \in H$ therefore $aH=bH$ ?2017-07-04
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    @JoaquinBrandan yes, this is exactly what I wrote at the very beginning2017-07-05
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    @AndreasT thanks. you added a bunch of inclusions that kind of went over my head. That's why i ask.2017-07-05