# Proving that the equivalence classes are the left and the right cosets and are well-defined and bijective

My question is:

The equivalence classes are the left and the right cosets of $$H$$ in $$G$$, respectively. Prove that the function $$\varphi : G/\sim_H \to G/\rho_H$$ given by $$\varphi (xH) = H(x^{−1})$$ is well-defined and bijective.

I am having a hard time understanding how to approach this problem. I know that because of Lagrange's Theorem there will be the same number of elements in the right and left cosets. I am not sure how to use this information knowing that the equivalence class are the left and right cosets of a subgroup $$H$$ in $$G$$.

If the equivalence classes are written as: $$x\sim_H y\iff x^{−1}y \in H$$ and $$x\rho_H y\iff xy^{−1}\in H$$, I am not sure how that would imply that the function $$\varphi(xH) = Hx^{−1}$$, is well-defined and bijective?

Any type of hint or observation would be very appreciated!

• What does it mean, $Hx-1$? – Arnaud Mortier Oct 29 '19 at 21:11
• I forgot to add parentheses: H(x^-1) or H times the inverse of x – emitsch Oct 29 '19 at 21:13
• – Shaun Oct 29 '19 at 21:28
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun Oct 29 '19 at 22:27

You have to understand conceptually what it means for this map to be well-defined. It means, observe that we have defined it by mapping $$xH$$ to something that depends on $$x$$. But $$xH$$ is also equal to $$yH$$ for some $$y$$'s, what if we apply the same process to $$y$$ instead? Do we get the same result? If not, then your process is ill-defined.