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Let $G$ be a finitely presented groups defined by

$$G=\{x_1,\ldots,x_n\mid R_1(x_1,\ldots,x_n)=\cdots=R_m(x_1,\ldots,x_n)=1 \}.$$

Let this presentation be denoted by $P$.

Let $S_n$ be the symmetric group on $\{1,\cdots,n\}$. Define

$G_*(P)=\{x_1,\ldots,x_n\mid \bigcup_{\sigma\in S_n} R_1(x_{\sigma(1)},\ldots,x_{\sigma(n)})=\cdots= R_m(x_{\sigma(1)},\ldots,x_{\sigma(n)})=1\}$.

For example, if $G=\{x,y\mid x^2=y^4=xyxy=1\}$, then $$G_*(P)=\{x,y\mid x^2=y^4=xyxy=x^4=y^2=yxyx=1\},$$ which is isomorphic to the Klein four group.

My question:

It seems natural that people has studied this before. What is the name for this "symmetrising the relations" process and where can I find the references?

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    $\begingroup$ These remind me of symmetrically presented groups. $\endgroup$ – Shaun Mar 9 at 17:14
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    $\begingroup$ Your notation is prone to confusion: $G_*$ depends on the choice of presentation, and not only on $G$. There might be good reasons that this has not or seldom been studied, if this is not motivated by topology or geometry, and if it leads to nothing significant. Of course, it's not a reason not to have a look. $\endgroup$ – YCor Mar 9 at 17:51
  • $\begingroup$ A basic remark that the resulting groups are precisely the finitely presented groups having a generating $k$-element subset $S$ such that the group of all permutations $\mathfrak{S}(S)$ of $S$ extends to a finite group of automorphisms of $G$. I don't know if anything particularly significant can be said about this class of groups. $\endgroup$ – YCor Mar 9 at 17:56
  • $\begingroup$ @YCor, thank you! I have edited my question. $\endgroup$ – Zuriel Mar 9 at 17:59

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