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How does one find the rank of a normal closure of a subset?

In particular, I am studying this group: Let $K_n$ be the group with $n$ generators $x_1,\cdots,x_n$ satisfying the relation that each $x_i$ commutes with all its conjugates, where $i=1,\cdots,n$. Thus there is a group homomorphism $f_{n}:K_n\to K_{n-1}$ mapping $x_i$ to $x_i$ when $i\leq n$ and maps $x_n$ to the identity element. Consider the kernel of $f$: it is the conjugate closure of the one-element subset $\{x_n\}$ of $K_n$. Then what is the rank (I mean the minimum number of generators) of the normal closure of $\{x_n\}$?

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  • $\begingroup$ Is conjugate closure the same as normal closure? $\endgroup$
    – Nick Gill
    Commented Jul 4, 2013 at 8:58
  • $\begingroup$ Yes, they are exactly the same. $\endgroup$
    – Zuriel
    Commented Jul 4, 2013 at 11:16
  • $\begingroup$ As stated the rank of the normal closure of $\{ x_n \}$ in $K_n$ could be 1 if $K_n$ is abelian and infinite if $K_n$ is the free group. Do you also mean that $x_i$ only commutes with elements it is conjugate to? $\endgroup$
    – Neil Hoffman
    Commented Jul 10, 2013 at 20:47
  • $\begingroup$ @NeilHoffman, Yes, $K_n$ is neither Abelian (unless $n=1$) nor the free group. In $K_n$, each generator $x_i$ commutes with all the conjugates of $x_n$. When $n=2$, it is the discrete Heisenberg group. $\endgroup$
    – Zuriel
    Commented Jul 15, 2013 at 4:18

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