Query regarding a Lemma from a paper.

Question

This is from a paper. Let me introduce the required definitions and post my query:
Definition: A vector $\mathbf{x}$ is a $q$-dimensional vector of probabilities defined as $\mathbf{x}=(x_0,x_1,\dots ,x_{q-1})$; $\forall{i},x_i\geq0$; $\sum_{i=0}^{q-1}x_i=1$ . The indices $0,1,\dots,q-1$ are to be interpreted as elements from the finitie field $GF(q)$.
Definition: An operation $\mathbf{x}^{+g}\triangleq (x_g,x_{1+g},\dots,x_{(q-1)+g})$ for any element $g \in GF(q)$.
Definition: Another operation $\mathbf{x}^{\times g}\triangleq (x_0,x_{g},\dots,x_{(q-1).g})$ for any element $g \in GF(q)$.
The operations ${+g}$ and $\times g$ are reversible with analogous definitions $(\mathbf{x}^{+g})^{-g}=\mathbf{x}$ and similarly for $g \neq 0$, $(\mathbf{x}^{\times g})^{g^{-1}}=\mathbf{x}$.

Question: The question I have is about the following Lemma that the author states:
Lemma: For $g \in GF(q)\setminus \{0\}$ and $i \in GF(q)$, $(x^{+i})^{\times g}=(x^{\times g})^{+ (i.g^{-1})}$.
The proof, as the author says, is to be obtained by looking at a specific index $k$ on the 2 sides of the equation but I'm unable to prove it

My interpretation:
Take an element $x_k,k \in (0,1,\dots,q-1)$.
LHS: The element $x_k$ becomes $x_{(k+i).g}$
RHS: The element $x_k$ becomes $x_{k.g+i.g^{-1}}$
Are the above two equal?

Answer 1

I think that you have misunderstood the definition of the two operations. The vectors $\mathbf{x}$ here are just functions from $GF(q)$ to the reals. Let us define addition mappings $$ A_i:GF(q)\to GF(q), k\mapsto k+i $$ for all $i\in GF(q)$, and multiplication mappings $$ M_g:GF(q)\to GF(q), k\mapsto kg $$ for all $g\in GF(q), g\neq 0$. The definitions on the vectors then mean $$ \mathbf{x}^{+i}:=\mathbf{x}\circ A_i, $$ and $$ \mathbf{x}^{\times g}:=\mathbf{x}\circ M_g. $$ Therefore $$ (\mathbf{x}^{+i})^{\times g}=(\mathbf{x}\circ A_i)\circ M_g $$ and $$ (\mathbf{x}^{\times g})^{+(i g^{-1})}=(\mathbf{x}\circ M_g)\circ A_{ig^{-1}}. $$ But $$ (A_i\circ M_g)(k)=A_i(kg)=kg+i=(k+ig^{-1})g=M_g(k+ig^{-1})=M_g(A_{ig^{-1}}(k)) $$ for all $k\in GF(q)$, so $A_i\circ M_g=M_g\circ A_{ig^{-1}}.$

The claim follows from this.

Another way of looking at it is the following. The component of $(\mathbf{x}^{+i})^{\times g}$ at position $k$ is the component of $\mathbf{x}^{+i}$ at position $kg$, which is the component of $\mathbf{x}$ at position $kg+i$.

Similarly the component of $(\mathbf{x}^{\times g})^{+(i g^{-1})}$ at position $k$ is the component of $\mathbf{x}^{\times g}$ at position $k+ig^{-1}$, which is also the component of $\mathbf{x}$ at position $kg+i$.

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It is very easy to go astray here, been there :-), so I add an example. If $q=p$, then the field operations are just modular arithmetic. Then $$ (x_0,x_1,\ldots,x_{p-1})^{\times (-1)}=(x_0,x_{p-1},x_{p-2},\ldots,x_1), $$ so $$ \left((x_0,x_1,\ldots,x_{p-1})^{\times (-1)}\right)^{+1}=(x_0,x_{p-1},x_{p-2},\ldots,x_1)^{+1}=(x_{p-1},x_{p-2},\ldots,x_1,x_0). $$ because the $^{+1}$ operation simply rotates the components one position to the left.

As predicted, we get the same result from the calculations (here $g=-1, i=1$, so $ig^{-1}=-1$) $$ (x_0,x_1,\ldots,x_{p-1})^{+ (-1)}=(x_{p-1},x_0,x_1,\ldots,x_{p-2}) $$ and $$ \left((x_0,x_1,\ldots,x_{p-1})^{+ (-1)}\right)^{\times(-1)} =(x_{p-1},x_0,x_1,\ldots,x_{p-2})^{\times(-1)} =(x_{p-1},x_{p-2},\ldots,x_1,x_0). $$ All because here $^{+1}$ operation simply rotates the components one position to the left, $^{+(-1)}$ rotates the components one position to the right, and $^{\times(-1)}$ reverses the order of the other components save the first.