Let $H$ be a subgroup of a group $G$. Show that for $a,b\in G$ we have $aH=bH$ if and only if $a^{-1}b\in H$

Question

Group $H \leq G$. Then for $a,b\in G$, $aH=bH \iff a^{-1}b\in H$

My Proof:

($\Rightarrow$) Suppose $aH = bH$, then

$a^{-1}aH=a^{-1}bH \ldots e = ab^{-1} \in H$

How to I prove conversely ($\Leftarrow$)?

say suppose $ab^{-1} \in H$...

Answer 1

First a small observation: Let $t\in H$ then $tH = H$. Proof:

1. We show $tH \subseteq H$. For every $x\in H$ we know that $tx \in H$ (because H is a (sub)group). This means that all elements of $tH$ are actually elements of $H$.
2. We show that $H \subseteq tH$. For all $x\in H$ we know that $t^{-1}x \in H$ (because H is a (sub)group). We can use the identity $x=tt^{-1}x$. Since $t^{-1}x \in H$ it follows that $tt^{-1}x\in tH$. Thus $x$ is both in $H$ and in $tH$, thus $H \subseteq tH$.
3. Thus $tH = H$.

Now to your problem:

1. Let $a^{-1}b \in H$. Thus we have $H = a^{-1}bH$. Thus $aH = a (a^{-1}b H) = (aa^{-1}b)H = bH$.
2. Let $aH = bH$. Thus there is a $t\in H$ such that $at = be$, where $e$ is the neutral element of $H$. Thus $t=a^{-1}b$. Thus we conclude $a^{-1}b\in H$.

Answer 2

Suppose $a^{-1}b\in H$. Then $a^{-1}b=h$, some $h\in H$, so now $b=ah\in aH$ and since $b=be\in bH$, therefore $aH\cap bH\neq\emptyset$. Hence $aH=bH$.

Answer 3

For the converse, assume $a^{-1}b\in H$, we want to show $aH = bH$.

Let $a^{-1}b = h$ for $h\in H$.

1. Suppose $x\in aH$. Let $x=ah_1$ for $h_1\in H$. $x^{-1}=h_1^{-1}a^{-1}\implies x^{-1}b=h_1^{-1}(a^{-1}b) = h_1^{-1}h\implies x = bh^{-1}h_1\in bH$. This shows $aH\subseteq bH.$
2. $x\in bH \implies a^{-1}x\in a^{-1}bH = hH = H\implies x\in aH$. This shows $bH\subseteq aH.$

Hence $aH=bH$.