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I'm attempting to understand why two cosets of a subgroup $U \leq G$, $aU$ and $bU$, must be disjoint in the proof of Lagrange's theorem. Here is my sketch so far of a proof by contradiction.

Suppose $aU \cap bU \neq \emptyset$. Then $\exists u \in U : a = bu$. Then $a \cdot b^{-1} \in U$.

Assuming $a$ and $b$ are 2 distinct elements of $G$ not in $U$, why can't this be true? Thanks.

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    $\begingroup$ Have you checked this post? $\endgroup$ – Couchy311 Jul 26 '17 at 5:31
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You have misinterpreted the statement. There are $[G\colon U]$ distinct cosets of $U$. As you said, if $a\neq b$ and $aU\cap bU\neq \emptyset$, then you must have $b^{-1}a\in U$ (I think you flipped the order). In fact, you can show that either $aU\cap bU=\emptyset$, or $aU=bU$.

If $aU\cap bU\neq\emptyset$, then $b^{-1}a\in U$ and $a^{-1}b\in U$, since $U$ is a subgroup. For any element $bu_0\in bU$, you can write $bu_0$ as $a(a^{-1}bu_0)$, which is in $aU$. The other direction is similar. Thus $aU=bU$.

Of course if $aU\cap bU=\emptyset$ then we are done.

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We prove the fact that " Any two left (right) cosets of a subgroup $U$ of a group $G$ is either $disjoint$ OR $identical$ ".

Assume that two left cosets $aU$ and $bU$ are NOT disjoint. Then there is an element $h\in G$ such that $h\in aU,bU$. Say, $h=au_1$ and simultaneously $h=bu_2$. Then we get $a=bu_2u_1^{-1}$. Let $au\in aU$ be any arbitrary element. Then $au=bu_2u_1^{-1}u\in bU$ (Since $U$ is a subgroup of $G$, $u_2u_1^{-1}u\in U$). Thus $aU \subset bU$. Mimicking the above argument, one can show that $bU \subset aU$. Hence $aU=bU$.

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Two distinct $a$ and $b$ in $G\setminus U$ don't necessarily give different cosets $aU$ and $bU$.

What you must show from your assumption $aU\cap bU \neq \emptyset$ is that $aU=bU$.

To show this: From $a=bu$ you get $u=b^{-1}a$ (multiply from the left) so $b^{-1}a\in U$, also it's inverse $a^{-1}b \in U$. So for any $bu_1 \in bU$ we have that $bu_1 = (aa^{-1})bu_1 = a(a^{-1}b)u_1 \in aU$, since $(a^{-1}b)u_1\in U$. Hence $bU \subset aU$.
The other direction $aU\subset bU$ can be shown similarly (use that $b^{-1}a\in U$).

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