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Left cosets, not right cosets!
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Jeremy Rickard
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Let $G=C_2\ast C_3$ be the free product of cyclccyclic groups $C_2=\langle a\rangle$ and $C_3=\langle b\rangle$.

Let $A$ be the subset of $G$ consisting of $$b,bab,babab,\dots$$ together with all reduced words ending with $a$ except for $$ba,baba,bababa,\dots.$$

Then $A$ is a set of rightleft coset representatives for both $B_1=\langle a\rangle$ and $B_2=\langle b\rangle$, so $G=A\cdot B_1=A\cdot B_2$.

Let $G=C_2\ast C_3$ be the free product of cyclc groups $C_2=\langle a\rangle$ and $C_3=\langle b\rangle$.

Let $A$ be the subset of $G$ consisting of $$b,bab,babab,\dots$$ together with all reduced words ending with $a$ except for $$ba,baba,bababa,\dots.$$

Then $A$ is a set of right coset representatives for both $B_1=\langle a\rangle$ and $B_2=\langle b\rangle$, so $G=A\cdot B_1=A\cdot B_2$.

Let $G=C_2\ast C_3$ be the free product of cyclic groups $C_2=\langle a\rangle$ and $C_3=\langle b\rangle$.

Let $A$ be the subset of $G$ consisting of $$b,bab,babab,\dots$$ together with all reduced words ending with $a$ except for $$ba,baba,bababa,\dots.$$

Then $A$ is a set of left coset representatives for both $B_1=\langle a\rangle$ and $B_2=\langle b\rangle$, so $G=A\cdot B_1=A\cdot B_2$.

Source Link
Jeremy Rickard
  • 35.2k
  • 2
  • 110
  • 151

Let $G=C_2\ast C_3$ be the free product of cyclc groups $C_2=\langle a\rangle$ and $C_3=\langle b\rangle$.

Let $A$ be the subset of $G$ consisting of $$b,bab,babab,\dots$$ together with all reduced words ending with $a$ except for $$ba,baba,bababa,\dots.$$

Then $A$ is a set of right coset representatives for both $B_1=\langle a\rangle$ and $B_2=\langle b\rangle$, so $G=A\cdot B_1=A\cdot B_2$.