Let $G$ be a finite group $d(G) = \min_{<S>=G}|S|$. Suppose that $|X|=|Y|=d(G)$ and $<X>=<Y>=G$. Let $|g|_X$ be the word length of $g$ with respect to $X$ and $|g|_Y$ be the word length of $g$ with respect to $Y$.
Suppose that $G=\{1,g_2,\cdots,g_n\}$ is not the trivial group. Does there exist a permutation $\pi \in S_n$ such that:
$$ | g_{\pi(i)}|_X = |g_{i}|_Y$$
for all $i=1,\ldots,n$?
In other words, if I consider for example the sum:
$$\sum_{g \in G} f(|g|_X)$$
then this sum should be equal, if the question, can be answered with yes, to
$$\sum_{g \in G} f(|g|_Y)$$
where $f : \mathbb{N_0} \rightarrow \mathbb{R}$ is any function, since we run through all elements $g$ of $G$ and because there (should) exist this permutation $\pi$ permuting the elements of $G$ and preserving the word length.
This might be a trivial question or worse, it could be wrong, but I could not find anything in books or on the internet related to it.
Thanks for your help!
Also, I am not sure if I have properly tagged the question, so that people interested in this kind of question can find it. So if you know of better tags, feel free to change this.
Interesting question ! The answer is no and we take $G =S_4$, $X = (s_1,s_2,s_3) $ and $Y = (s_1, s_1s_2, s_1s_3)$. Here $s_i = (i,i+1)$ with the usual relations $s_is_j = s_js_i$ if $|i - j|>1$ and $s_is_js_i = s_js_is_j$ if $i = j \pm 1$.
We have $\{g \in S_4 : d_X(g) = 2\} = \{s_1s_2, s_1s_3, s_2s_1, s_2s_3, s_3s_2\}$ ($5$ elements).
On the other hand $\{g \in S_4 : d_Y(g) = 2\} = \{ s_2s_1, s_2, s_3, s_1s_2s_1, s_1s_2s_1s_3, s_3s_2 \}$ ($6$ elements).
More precisely, if we call $t_1 = s_1, t_2 = s_1s_2, t_3 = s_1s_3$ then :
$t_2^2 = s_1s_2s_1s_2 = s_2s_1$, $t_1t_2 = s_2, t_1t_3 = s_3, t_2t_1 = s_1s_2s_1, t_2t_3 = s_1s_2s_1s_3$ and $t_3t_2 = s_3s_2$.