Transfer homomorphism: is this proof correct?

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I'm doing a guided exercise to learn the transfer homomorphism. The preliminaries are these:

Given a finite group $G$ and a subgroup $H<G$, let $\{ x_iH \}_{i\in I}$ be the set of left cosets of $G$ with respect to $H$ once chosen the representatives $x_i$.

Now the exercise is divided in three steps (now I write only two of them):

i)Prove that, given $g \in G$ there exists uniques $j \in I$ and $h_i \in H$ such that $gx_i = x_jh_i$

ii)Show that $\prod_{i \in I} h_i $ is a well defined element of $\frac{H}{[H,H]} $ which doesn't depend on the choice of class representatives but only on the element $g$

I have done the first one and I have a proof for the second, but I don't know if is correct or not. This is the proof of ii)

Let $(x_1,...,x_t)$ and $(y_1,...,y_t)$ be two sets of class representatives, then there exists a permutation $\sigma \in S_t$ such that $\forall i=1,...t, \ \ x_iH=y_{\sigma(i)}H$.

Now, I have that given $g \in G$ $gx_i=x_jh_i \ \iff h_i=x^{-1}_jgx_i$ and we know that

$x^{-1}_jH=y^{-1}_{\sigma(j)}H$ and $x_iH=y_{\sigma(i)}H$

so if we compute $h_i \in \frac{H}{[H,H]}$ we have

$x^{-1}_jHgx_iH=y^{-1}_{\sigma(j)}Hgy_{\sigma(i)}H$

but the quotient between a group and the derived group is abelian and so

$x^{-1}_jgx_iH=y^{-1}_{\sigma(j)}gy_{\sigma(i)}H$ and $H \subset [H,H]$

so the classes are equal.

Is this proof right or not? Do you have some suggestion?