Why must cosets of a subgroup be disjoint in the proof of Lagrange's Theorem? 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.

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.
