Cosets of a subgroup do not overlap

This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.

This problem is from assignment $5$.

Prove directly that distinct cosets do not overlap.

Let $H$ be a subgroup of $G$ and let $a$, $b$, and $c$ be elements of $G$ such that $b\not\in aH$ and $c$ is in both $aH$ and $bH$. Then there are elements $h$ and $h^\prime$ in $H$ such that $c=ah$ and $c=bh^\prime$. So $ah=bh^\prime$ and $b=ahh^{\prime -1}$. But $hh^{\prime -1}\in H$ so $b\in aH$. This contradicts our original assumption. Therefore, there can be no element in more than one distinct coset.

Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.

Thanks.

To prove the result at hand, you need to start with two distinct cosets sharing a common element, and prove that the cosets coincide. In saying that, "let $a$, $b$, and $c$ be elements of $G$ such that $b \not\in aH$ and $c$ is in both $aH$ and $bH$", you are assuming that $aH$ and $bH$ are distinct cosets with non-empty intersection and that doesn't mean that $b \not \in aH$ [as of now]. Also, $b \not \in aH$ does not also mean that, $bH \neq aH$ [as of now].

So, I think your proof is faulty here.

I'd fix it this way.

Suppose $c$ belongs to distinct cosets, $aH$ and to $bH$, say $$c = ah_1 = bh_2$$ where $h_1$, $h_2 \in H$. Then $a = bh_2h_1^{-1}$. Any element of $bH$ has the form $bh$ for some $h \in H$ and, $$ah = b(h_2h_1^{-1}h) \in bH$$

Since $h$ was arbitrary in $H$, we see $aH\subseteq bH$ .

The reverse inclusion, $bH \subseteq aH$, follows by a similar argument (using the equation $b = ah_1h_2^{-1}$ instead and arguing similarly).

This proves the claim.

There is a round about way of doing this, which I'll nevertheless mention here: The approach through equivalence relations.

Let $H \subseteq G$ be a subgroup of $G$. Define $\sim$ on $G$ by, $a\sim b$ iff there exists $h \in H$ such that, $a=bh$ for $a,b \in G$. Prove that your notion of cosets coincide with the equivalent classes of $\sim$. Now, you know that distinct equivalence classes are disjoint and hence your result.

Note that this leads you into Lagrange's Theorem in finite groups and in fact, the following is also true.

Let $H \subset G$ such that for all $a,b \in G$, $aH \cap bH=\emptyset$ or $aH=bH$ holds. Then $\exists g \in G$ such that $gH$ is a subgroup of $G$.

In case you have difficulty proving the above result, please let me know.

Hope this helps.

Edited to add @Dylan's viewpoint of this problem from group actions: Note that the equivalence relation we have defined can be viewed as a group action by looking at the definition of orbits. (i.e.) search for the action whose orbits are the equivalence classes of $\sim$. You'll identify that this action is the left multiplication action.