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?

  • $\begingroup$ "for some $h \in H$ we find $bh = b$". You can be more specific than this. What element $h$ satisfies $bh = b$? $\endgroup$
    – user169852
    Apr 29 '20 at 4:53
  • 1
    $\begingroup$ @Bungo true, it must be that $h=e$ the identity element of $G$. And since $H$ is a subgroup of $G$, $e \in H$. $\endgroup$
    – Max
    Apr 29 '20 at 4:55
  • $\begingroup$ 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 $bH$. $\endgroup$
    – user169852
    Apr 29 '20 at 4:57
  • $\begingroup$ Correct, $e \in H$, hence $b = be \in bH = aH$. $\endgroup$
    – 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 $a$?

  • $\begingroup$ Is it possible to show the iff is true without using equivalence relations? $\endgroup$
    – Max
    Apr 29 '20 at 5:00
  • $\begingroup$ @max Yes, it is possible. See my edits above. $\endgroup$
    – Mnifldz
    Apr 29 '20 at 5:02
  • $\begingroup$ 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? $\endgroup$
    – Max
    Apr 29 '20 at 5:09
  • $\begingroup$ Assuming I'm not way off: Then $a \in bH$ implies $aH \subseteq bH$ and since $bH \subseteq aH$, this would imply $aH = bH$. $\endgroup$
    – Max
    Apr 29 '20 at 5:13
  • $\begingroup$ @Max Yes, you got it. $\endgroup$
    – Mnifldz
    Apr 29 '20 at 6:33

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