Properties of cosets; $aH = bH \iff b^{-1}a \in H$.

Question

I've found the concept of cosets to be strange when I've encountered them. I want to make sure that I'm understanding how to work with them.

Claim: Given a subgroup $H \leq G$ and $a,b \in G$, we have $aH = bH \iff b^{-1}a \in H$.

Attempt: Assume first that $aH = bH$. This means that as sets $$ \{g \in G \mid g = ah, h \in H\} = \{g \in G \mid g = bh, h \in H\}. $$ Therefore if $aH = bH$ then $ah_1 = bh_2$ for some $h_1,h_2 \in H$. This implies that $b^{-1}a = h_2h_1^{-1} \in H$. Conversely assume that $b^{-1}a \in H$. This means that there exists some $h \in H$ such that $b^{-1}a =h$. Then it follows that $a = bh \in bH$ and that $b = ah^{-1} \in aH$. If we pick any arbitrary $h' \in H$ we must have that $ah' = b(hh') \in bH$ but this means that $aH \subseteq bH$. Identically $bh' = a(h^{-1}h') \in aH$ so $bH \subseteq aH$ and we have $aH = bH$.

Is this overkill? Whenever I read problems or questions the manipulations on cosets seem to be much faster. How can I stop thinking of them as sets, or the quotient group $G/H$ as a "set of sets"?

Thanks in advance.

Answer 1

The proof is correct, and not necessarily overkill but I'm going to give some feedback on how you wrote it, which might tighten it up.

1. "Therefore if $aH=bH$ then $ah_1=bh_2$ for some $h_1,h_2\in H$." This is not very clear: it looks like a statement involving an existential statement on two elements. What you actually have is that for every $h_1\in H$ there exists $h_2\in H$ such that $ah_1 = bh_2$, and likewise for every $h_3\in H$ there exists $h_4\in H$ such that $ah_4=bh_3$. You can then deduce your statement, but one usually first picks an (arbitrary?) element from $H$ to get the thing started, rather than existentially pick two elements.

2. Note that this allows you to simplify the argument ans make it look more like the converse: pick $h_1=e$, and deduce that there exists $h\in H$ such that $a=bh$; then $b^{-1}a=h\in H$ is a simpler calculation. (In particular, notice that the only element of $G$ that you know for sure is in $H$ is $e$, so the only instantiation of $h_1$ that you can be sure is valid is $h_1=e$).

3. Once you get that "then there exists $h$ such that $a=bh\in bH$, you can now invoke the fact that two cosets of $H$ are either disjoint or equal. Since $a\in aH\cap bH$, it follows that $aH=bH$ and you are done. No need for double inclusion. But if you want to do double inclusion, once you have that $b^{-1}a\in H$ implies $aH\subseteq bH$, you can then note that because $H$ is a subgroup, then $b^{-1}a\in H$ implies $a^{-1}b=(b^{-1}a)^{-1}\in H$, which implies that $bH\subseteq aH$, giving equality. This exploits the fact that the roles of $a$ and $b$ are symmetric.

An alternative proof that uses the cosets is to note that $gH=H$ if and only if $g\in H$, which is easy to verify (just note that $e\in gH$ if and only if $g^{-1}\in H$ if and only if $g\in H$). So $$\begin{align*} aH = bH &\iff b^{-1}(aH) = b^{-1}(bH)\\ &\iff (b^{-1}a)H = (b^{-1}b)H\\ &\iff (b^{-1}a)H = eH\\ &\iff (b^{-1}a)H = H\\ &\iff b^{-1}a\in H. \end{align*}$$