I'm approaching the study of the transfer homomorphism between groups, and I found this exercise:
Given a finite group $G$ and a subgroup $H<G$, the set of left cosets of $H$ in $G$ is written $ \{x_iH \}_{i \in I}$ . Now I want to prove that, given $g \in G$ for all $i \in I$ there exists uniques $j\in I, h_i \in H$ such that $gx_i=x_jh_i$.
The fact that the indices of $x_i$ and of $h_i$ have to be equal for every $i \in I$ confuses me a little bit about the existence. In fact if I want to prove this for the first attempt I'm able to re-order the $h_i$ in order to make the equation hold (follows from the fact that cosets forms a partition of $G$), but once I've take this choice how can I show that I can find proper indices?
If I suppose the existence to be satisfied then the uniqueness is quite trivial because given $j,j' \in I$ then $gx_i=x_jh_i=x_{j'}h_i \implies x_j=x_{j'}$
Well, first think about what the question is really saying: if I take an element $g$ of my group $G$ and multiply it by a coset representative $x_i$,then it should land in another coset hence be written in the form $x_jh_i$ for some $h_i\in H$ and other coset representative $x_j$. (Since $G$ is fixed, $h_i$ only depends on $x_i$ hence the same indices, but a priori $x_j$ can be any representative hence the different index.) As you note, this follows from the fact that cosets partition $G$ (assuming you are allowed to use this fact for the exercise).