1. Oct 30, 2014

### Bashyboy

Here is a link to a proof which I am trying to understand.

http://groupprops.subwiki.org/wiki/Left_cosets_partition_a_group

The claim I am referring to is number 4, which is

Any two left cosets of a subgroup either do not intersect, or are equal.

Assuming that I am skeptical, then for all I know there are three cases: (1) $aH \cap bH \ne \emptyset$ but $aH \ne bH$; (2) $aH = bH$; or (3) $aH \cap bH = \emptyset$.

The truthfulness of (2) and (3) is made reasonable by simple examples and calculations involving certain groups, such as $D_4$. However, working with these examples, it is not clear that (1) is true or false; consequently, it remains as a possibility.

In the proof given in the link, they start out by supposing that $aH$ and $bH$ are not disjoint, that they could have some elements in common. Continuing on in the proof, we see that by supposing this is true, this undoubtedly leads to the cosets being equal; in doing this, they also show that it is not possible for $aH \cap bH \ne \emptyset$ but $aH \ne bH$.

But they don't treat whether it is possible for $aH \cap bH = \emptyset$ to be true. Why is that?

2. Oct 31, 2014

### jbunniii

It is certainly possible for $aH \cap bH = \emptyset$ to be true. This occurs if and only if $a$ and $b$ are in different cosets of $H$.

3. Nov 1, 2014

### Fredrik

Staff Emeritus
Because it's obvious that if they're disjoint, they're not equal. (This is true for any two non-empty sets, and we have $a\in aH$, $b\in bH$).

The goal is to prove that $aH$ and $bH$ are either disjoint or equal. They are clearly either disjoint or not disjoint, so it's sufficient to prove that if they're not disjoint, they're equal.

4. Nov 1, 2014

### Bashyboy

But how do I know if it is possible that they are disjoint? I understand that a disjunctive statement $p \vee q$ is true when at least one of the simple statements is true.

Is the only way to know that it is possible for two cosets to be disjoint is to work with simple examples, as I have already done?

5. Nov 1, 2014

### Fredrik

Staff Emeritus
The theorem holds for all groups $G,H$ such that $H$ is a subgroup of $G$. The proof doesn't rely on the existence of $a,b\in G$ such that $aH$ and $bH$ are disjoint. So the proof even holds for the case $H=G$.

Suppose that no such $a,b$ exist. Then $aH=H$ for all $a\in G$. (This follows from the theorem and the fact that $eH=H$). Let $a\in G$ be arbitrary. We have $a=ae\in aH=H$. So $G\subseteq H$. Since $H\subseteq G$ by assumption, this implies that $H=G$.

6. Nov 1, 2014

### jbunniii

If $H$ is a proper subgroup of $G$, then take any element $a \in G$ which is not in $H$. Which coset of $H$ contains $a$? Certainly not $H$ itself. Therefore $aH$, the coset containing $a$, is not the same as $H$. By the theorem, this forces $aH \cap H = \emptyset$.