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Problem from Artin's Algebra:

Prove directly that distinct cosets do not overlap.

I found the solution here.

But my question is different. The question itself says distinct cosets. So if their is at least, one element in common, they can not be distinct. On the other hand if distinct implies "two cosets $aH$ and $bH$, then $a\neq b$"(which is not obviously), then that does not imply $aH\neq bH$, or at least they do not overlap.

Two distinct cosets creates two different equivalence class. So their is no question of overlapping obviously.

I could not understand the question.

Any help would be appreciable.

Note: Since this question may get marked by duplicate. You don't have to flag. Just comment duplicate. I will delete the post. Btw my question is not to answer of the problem. So hope not getting marked by duplicate.

Community
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MAN-MADE
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    Two sets are said to be distinct when they are not equal (not when they have no element in common). When two sets have no element in common, they are said to be disjoint. –  Jul 13 '17 at 04:31
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    Cosets are either disjoint or equal. So if two cosets have at least one element in common, they are essentially the same. And yup, if $a\neq b$, it can be the case that $aH=bH$ since $a$ and $b$ can be distinct representatives of the same coset. –  Jul 13 '17 at 04:31
  • You can probably use the fact that you have an equivalence relation to show that if two cosets shared a common element, you could use transitivity using that element to show that every element of the first is contained in the other, and vice versa. – Arkady Jul 13 '17 at 04:32
  • "Distinct" does not mean they don't overlap. "Disjoint" means no overlap. "Distinct" just means they are not identical. "Distinct" and "disjoint" are two different words. – bof Jul 13 '17 at 04:41

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Two sets are distinct if one has an element that isn't the element of the other. So the question is to prove that if cosets $aH$ and $bH$ are distinct, then $aH\cap bH=\emptyset$. But it's easier to prove the equivalent statement, that if $aH$ and $bH$ have an element in common, then $aH=bH$.

Assume $aH\cap bH\ne\emptyset$ and let $c\in aH\cap bH$. An important property of cosets (can you prove this?) is that if $c\in aH$ then $cH=aH$. (Then also $cH=bH$ etc....)

Angina Seng
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This is to say that for distinct elements $a,b \in G$, $aH=bH \iff aH \cap bH \neq \emptyset$. In other words, if they "coincide" so that they share an element in common, they were in fact the same coset.

Andres Mejia
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