Consider a topological group $\mathrm{D},$ whose group operation I shall denote multiplicatively. The following result must be true but I fail to find a proof of it (except in the commutative case, which is not what I need). Consider a permutation $\sigma$ of $1, \ldots, p.$
Is the mapping $\mathrm{D}^p \to \mathrm{D}$ given by $(x_1, \ldots, x_p) \mapsto x_{\sigma(1)} \cdots x_{\sigma(p)}$ continuous with respect to the product topology on $\mathrm{D}^p$ induced by that of $\mathrm{D}$?
Again, in the commutative case $\sigma$ has no effect, thus one can write $$(x_1 \cdots x_p) (y_1 \cdots y_p)^{-1} = (x_1 y_1^{-1}) \cdots (x_p y_p^{-1});$$ one can consider then a neighbourhood $\mathrm{T}$ of the neutral element and another neighbourhood $\mathrm{K}$ (of the neutral element again) such that $\underbrace{\mathrm{K} \cdots \mathrm{K}}_{n \text{ times}} \subset \mathrm{T}$, then it suffices to take $x_i \in y_i \mathrm{K}$ for $1 \leq i \leq p$. I ask now, can this proof be modified slightly to allocate the noncommutative case?
The permutation is largely irrelevant; your map is just the composition of the permutation-of-coordinates map $D^p\to D^p$ (which is a homeomorphism) with the multiplication map $\mu_p:(x_1,\ldots,x_p)\to x_1\cdots x_p$.
The continuity of $\mu_p$ is an induction on $p$. For $p=2$ this is part of the definition of topological group. In general if $\mu_p$ is continuous, so is $(x_1,\ldots,x_{p+1})\mapsto x_1\cdots x_p$. Clearly $(x_1,\ldots,x_{p+1})\mapsto x_{p+1}$ is continuous. From the continuity of multiplication, $\mu_{p+1}:(x_1,\ldots,x_{p+1})\to (x_1\cdots x_p)x_{p+1}$ is continuous.