Reindexing the composition of two families of elements of a group

Question

Let $n,m\in\mathbb{N}_{>0}$ and let $(x_{i})_{i\in[1,m]}$ and $(y_{j})_{j\in[1,n]}$ be two families of elements of a group $G$ whose law is written multiplicatively. How can I rewrite the composition $$\prod_{i=1}^{m} x_{i}\cdot(\prod_{j=1}^{n} y_{j})^{-1}$$ as a family of elements of $G$ over the interval $[1,m+n]$? I know that for $$\prod_{i=1}^{m} x_{i}\cdot(\prod_{j=1}^{n} y_{j})$$ I have $(z_{k})_{k\in[1,m+n]}$ such that $$\prod_{k=1}^{m+n} z_{k}=\prod_{i=1}^{m} x_{i}\cdot(\prod_{j=1}^{n} y_{j}),$$ where \begin{equation*} z_k= \begin{cases} x_k, & 1\leq k\leq m; \\ y_{k-m}, & m+1\leq k\leq m+n. \end{cases} \end{equation*}

I tried defining a function $u$ such that \begin{equation*} u_k= \begin{cases} x_k, & 1\leq k\leq m; \\ y_{(n+1)-(k-m)}, & m+1\leq k\leq m+n. \end{cases} \end{equation*} but I ran into a problem whilst trying to show that $$(\prod_{j=1}^{n} y_{j})^{-1}=\prod_{k=m+1}^{m+n} y_{(n+1)-(k-m)}^{-1};$$ specifically, for the inductive step (induction over $n$ with $m$ fixed) we have $$\prod_{k=m+1}^{m+(n+1)} y_{((n+1)+1)-(k-m)}^{-1}=\prod_{k=m+1}^{m+n} y_{((n+1)+1)-(k-m)}^{-1}\cdot y^{-1}_{1},$$ but the expression $\prod_{k=m+1}^{m+n} y_{((n+1)+1)-(k-m)}^{-1}$ is different from what's assumed for $n$: namely $\prod_{k=m+1}^{m+n} y_{(n+1)-(k-m)}^{-1}$.

Any hints will be appreciated.

Answer 1

At first glance, I'd do it step-by-step. If $m=n$ then

$$\prod_{i=1}^{m}x_i\left(\prod_{j=1}^{m}y_j\right)^{-1} = \prod_{k=1}^{m}x_ky_k^{-1}.$$

If $m\gt n$ then

$$\prod_{i=1}^{m}x_i\left(\prod_{j=1}^{n}y_j\right)^{-1} = \prod_{i=1}^{m-n}x_i\prod_{l=1}^{n}x_{l+m-n}\left(\prod_{j=1}^{n}y_j\right)^{-1} = \prod_{i=1}^{m-n}x_i\prod_{k=1}^{n}x_{k+m-n}y_k^{-1}.$$

If $m\lt n$ then

$$\prod_{i=1}^{m}x_i\left(\prod_{j=1}^{n}y_j\right)^{-1} = \prod_{i=1}^{m}x_i\left(\prod_{j=1}^{n-m}y_j\prod_{l=1}^{m}y_{l+n-m}\right)^{-1} = \prod_{j=1}^{n-m}y_j\prod_{k=1}^{m}x_ky_{k+n-m}^{-1}.$$

Finally, try to combine all chances in one $\prod_{k=1}^{m+n}z_{k}$.