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Let $F_n$ be the free group generated by $x_1,\cdots,x_n$ and for each $1\leq i\leq n$, let a homomorphism $d_i:F_n\to F_{n-1}$ be defined as follows:

  • $d_i(x_r)=x_r$, if $i>r$;
  • $d_i(x_r)=1$, if $i=r$;
  • $d_i(x_r)=x_{r-1}$, if $i<r$.

The equalizer $E_n\leq F_n$ defined by $$E_n=\{x\in F_n\mid d_1(x)=d_2(x)=\cdots=d_n(x)\}$$ is clearly a subgroup of $F_n$, and thus a free group (Every subgroup of a free group is free).

Question: What is a minimal generating set of $E_n$?

My guess is the set consists of the following elements may be a (Is it minimal?) generating set; but I am not sure if it is correct:

  1. $x_{\sigma(1)}x_{\sigma(2)}\cdots x_{\sigma(n)}$;

  2. $[x_{\sigma(1)},x_{\sigma(2)}][x_{\sigma(1)},x_{\sigma(3)}]\cdots[x_{\sigma(1)},x_{\sigma(n)}][x_{\sigma(2)},x_{\sigma(3)}]\cdots[x_{\sigma(2)},x_{\sigma(n)}]\cdots[x_{\sigma(n-1)},x_{\sigma(n)}]$;

  3. $[[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(3)}][[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(4)}]\cdots[[x_{\sigma(n-2)},x_{\sigma(n-1)}],x_{\sigma(n)}]$;

  4. $\cdots$;

  5. $[\cdots[[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(3)}],\cdots x_{\sigma(n)}]$,

where $[x,y]=x^{-1}y^{-1}xy$ and $\sigma\in\Sigma_n$, the symmetric group acts on $\{1,2,\cdots,n\}$.

EDIT: It was pointed out in the comment that the set I proposed is not a generating set; it is not even a subset of $E_n$ in general.

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    $\begingroup$ I don't think your elements are actually in $E_n$ unless $\sigma$ is either the identity or the order-reversing permutation. $\endgroup$
    – Eric Wofsey
    Commented Jan 19, 2015 at 8:34
  • $\begingroup$ @EricWofsey, you are exactly right!! If I restrict my $\sigma$ to the identity and order-reversing permutations, could the list be a generating set? $\endgroup$
    – Zuriel
    Commented Jan 19, 2015 at 8:38
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    $\begingroup$ Thanks @მამუკაჯიბლაძე for the reference! But I do not see how Wu's result answer my question. $\endgroup$
    – Zuriel
    Commented Jan 19, 2015 at 11:57
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    $\begingroup$ Wu has free generators for the kernel of $E_n\to F_{n-1}$; it is true that one still has to determine the image, but I think this also can be done using Wu's calculations. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 19, 2015 at 13:44
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    $\begingroup$ More about the image - it is the intersection of the diagonal $F_{n-1}\subset F_{n-1}^n$ with the image of $(d_1,...,d_n):F_n\to F_{n-1}^n$ and I believe one can compute it using known methods. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 19, 2015 at 13:53

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