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Oct 10, 2018 at 12:10 comment added Derek Holt Sorry $G$ should be $W$. I don't have a $G$. I have $W = A^d \rtimes P$. Of course $P=S_n$ in your question.
Oct 10, 2018 at 12:09 history edited Derek Holt CC BY-SA 4.0
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Oct 10, 2018 at 12:07 vote accept Chris Russell
Oct 10, 2018 at 11:29 comment added Chris Russell I think I get the idea of how it works now but I'm not sure what $G$ is in the second last paragraph of your updated answer, which is confusing me. In one place I'm interpreting that $A^d$ is a subgroup of $G$ but in another $g \in G$ where I thought $g$ is an element of $P$.
Oct 10, 2018 at 10:49 history edited Derek Holt CC BY-SA 4.0
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Oct 10, 2018 at 10:48 comment added Derek Holt My answer was incomplete because it only described the $A^d$-classes. Calculating the $G$-classes from thsi information is not difficult - I have added some further explanation.
Oct 10, 2018 at 9:32 comment added Chris Russell Thank you Derek, I see how an element $(g, x)$ is conjugate to something in the set $R$ you define. If I write $$R_g := \{\,(g,x) \mid x_i = 1 \ \mathrm{for}\ i \not\in \{r_1,\ldots,r_s\},\, x_i \in \{x^{(1)},\ldots,x^{(k)}\}\ \mathrm{for}\ i \in \{r_1,\ldots,r_s\}\,\}$$ is there a way to describe a subset of $$\bigcup_{g \in G}R_g$$ which contains only one representative from each conjugacy class? Also is there a way to determine how many elements are in a each conjugacy class?
Oct 9, 2018 at 18:12 history answered Derek Holt CC BY-SA 4.0