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I am tryng to prove the below result from Artin's algebra.

Prove directly that distinct cosets do not overlap.

The part of the problem that confuses me mostly is the word 'directly,' because the proofs I wrote down were in effect contradiction or contraposition. I am assuming (someone please correct me if I'm wrong) that 'distinct' means 'generated by different elements,' so at the very least I have $a \neq b$. In reality, it should mean $aH \neq bH$, but that's assuming the conclusion. Here is my attempt. It could also mean that, as sets, $aH \neq bH$ so there is something in $aH$ not in $bH$ or vice versa, but there could be an overlap. I'm not sure which definition to use a-priori.

Let $H \subset G$ a subgroup, and $aH, bH$ left cosets of $H$ in $G$. If $x \in aH \cap bH$, then there exist $h_1, h_2 \in H$ such that $x = ah_1 = bh_2$. In particular, $a = b(h_2 h_1^{-1})$, so given $y = ah \in aH$, we have $y = b\left(h_2 h_1^{-1} h\right)$, so $y \in bH$, so $aH \subset bH$. Similarly, $bH \subset aH$, so $aH = bH$, so the left and right costs are not distinct provided that they contain some overlap.

This isn't a direct proof, however. An alternative is to use the fact that cosets are defined an equivalence relation. In particular, $x \in aH$ provided that $a^{-1} x \in H$. Since cosets are defined by an equivalence relation, they are themselves equivalence classes, and equivalence classes are by definition a partition, so they are either identical or disjoint. If two cosets are "distinct," then they're not equal, so they must be disjoint. I'm not sure if this is the "direct" approach that Artin is asking for.

user861776
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2 Answers2

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If $x\in aH\cap bH$ there are $h_1, h_2\in H$ so $ah_1 = x = bh_2$. Then $a = bh_2h_1^{-1}$. Can you do the rest?

ncmathsadist
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  • I believe this is essentially the proof that I gave. The above implies $a \in bH$. Alternatively, we have $b = ah_1 h_2^{-1}$, so $b \in aH$. This in turn implies that $aH = bH$. But this isn't quite proving it 'directly.' – user861776 Mar 28 '21 at 23:52
  • @user861776 when you say you're looking for a "direct" proof, you mean you want to avoid a proof by contradiction, correct? – Matthew H. Mar 29 '21 at 00:00
  • Yes, or contraposition, which I believe is what I wrote in my attempt. – user861776 Mar 29 '21 at 00:02
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For a subgroup $H\le G$, the relation $a\sim b \iff ab^{-1}\in H $ defines an equivalence relation on $G $. As a result the cosets partition $G$.

  • I definitely had this in mind for a direct proof, and it will follow from the properties of equivalence classes. Is there another way, though, to show that $aH \neq bH$ implies $aH \cap bH = \emptyset$ without appealing to equivalence relations? – user861776 Mar 29 '21 at 04:33