Proving equivalence of statements about group cosets

I found the following statement with regards to group cosets:

Let $$H$$ be a subgroup of a group $$G$$ and suppose that $$a, b \in G$$.

If $$aH = bH$$, then $$b \in aH$$.

My attempt at writing a proof:

Assume $$aH = bH$$, then for some $$h \in H$$ we find $$b \circ h = b$$, thus $$b \in bH$$.

But $$bH = aH$$, therefore $$b \in aH$$ is also true.

My question: Is the reverse also true? Meaning, if $$b \in aH$$, then $$aH = bH$$? In other words, can we create an iff statement?

I tried to show that if $$b \in aH$$, then $$aH = bH$$, but didn't get very far.

EDIT: Is it possible to show the iff is true without using equivalence relations?

• "for some $h \in H$ we find $bh = b$". You can be more specific than this. What element $h$ satisfies $bh = b$?
– user169852
Apr 29 '20 at 4:53
• @Bungo true, it must be that $h=e$ the identity element of $G$. And since $H$ is a subgroup of $G$, $e \in H$.
– Max
Apr 29 '20 at 4:55
• Yes, the converse is also true. Every element of $G$ is contained in exactly one coset of $H$. Clearly $b \in bH$. If also $b \in aH$, then because there's only one coset containing $b$, this forces $aH = bH$. Thus the coset containing $b$ has (at least) two names: $aH$ and $bH$.
– user169852
Apr 29 '20 at 4:57
• Correct, $e \in H$, hence $b = be \in bH = aH$.
– user169852
Apr 29 '20 at 4:58

Because coset membership is an equivalence relation then cosets are either disjoint or identical, therefore the converse will also be true. Therefore an iff statement is possible.

Your proof is a little clunky. Notice that since $$aH = bH$$ then we have $$a^{-1}bH = H$$ hence we can write $$a^{-1}b = h \in H$$. What does this tell you about $$b$$?

For the converse, $$b \in aH$$ implies that $$bH \subseteq aH$$, since for any $$h, h' \in H$$ we have $$b = ah$$ and therefore $$bh' = ahh'$$, so $$bh' \in aH$$. What does this tell you about the element $$a$$?

• Is it possible to show the iff is true without using equivalence relations?
– Max
Apr 29 '20 at 5:00
• @max Yes, it is possible. See my edits above. Apr 29 '20 at 5:02
• Thanks for your update. So $b=ah$ makes sense, for some $h \in H$. Then $a = bh^{-1}$ and $a \in bH$. Is that right?
– Max
Apr 29 '20 at 5:09
• Assuming I'm not way off: Then $a \in bH$ implies $aH \subseteq bH$ and since $bH \subseteq aH$, this would imply $aH = bH$.
– Max
Apr 29 '20 at 5:13
• @Max Yes, you got it. Apr 29 '20 at 6:33