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Let $G$ be a group and $F_G$ the free group on the set $G$. Then there exists a canonical surjective morphism ${\rm can}: F_G \to G \to 1$ constructed as follows: let $(e_x)_{x \in G}$ be a copy as a set of $G$; then $F_G$ is the free group generated by $(e_x)_{x \in G}$ -- then ${\rm can}$ is the unique morphism of groups such that ${\rm can} (e_x) := x$, for all $x\in G$.

Is the set $\{e_x e_y e_{xy}^{-1}\}$ a system of generators of ${\rm Ker} ({\rm can})$?

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  • $\begingroup$ .@Gigel Militaru I put a remark in my answer but, now, am not sure of the real statement you are looking for. Can you elaborate a bit ? $\endgroup$ Apr 19, 2015 at 3:29
  • $\begingroup$ @Duchamp Gérard H. E., The question arise in the context of the classical theory of Schur concerning the covers and universal central extensions for (perfect) groups. I am trying to obtain a similar results for Hopf algebras. Unfortunately, I did not found a detailed proof (the one that can be exposed to students at blackboard :) ) of the original theorem of Schur: probably is buried somewhere in the literature. In this context the theorem reduce to ... the above question. :) $\endgroup$ Apr 19, 2015 at 6:10
  • $\begingroup$ .@GigelMilitaru I worked in symmetric functions and Hopf algebras some years ago, maybe I can help, what is this theorem by Schur ? (even if the form is incomplete and/or erroneous) $\endgroup$ Apr 19, 2015 at 10:25
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    $\begingroup$ @Duchamp Gérard H. E - Theorem (Schur): A group G has an universal central extension (UCS) iff G is a perfect group. An UCS of a group $G$ is a pair $(E, p)$, where $p: E \to G$ is an epimorphism such that ${\rm Ker} (p)$ is contained in the center of $E$ and $(E, p)$ is an intial object in the category of all these pairs (with the obvious morphisms). The above question is the key point of proving Schur theorem at the level of groups: now, we just have to take $E$ the derived group of the quotient of $F_G$ by the normal subgroup generated by all $[e_x e_y e_{xy}^{-1}, \, e_z]$. $\endgroup$ Apr 20, 2015 at 7:04

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The answer is finally yes.

Write $$f(x,y)=e_xe_ye_{xy}^{-1};\qquad R=\{f(x,y):x,y\in G\}\subset F_G.$$ Clearly the presentation $$\langle (e_g)_{g\in G}\mid (f(x,y))_{x,y\in G}\rangle$$ is a presentation of $G$, more precisely if $\Gamma$ is the group defined by this presentation and $\bar{e}_g$ the image of $e_g$ in $\Gamma$, we have a unique homomorphism $\phi:\Gamma\to G$ mapping $\bar{e}_g\mapsto g$ for all $g$. The first claim is that $\phi$ is an isomorphism.

Proof: [added in edit] clearly $\phi$ is surjective. Now in $\Gamma$, the inverse of $\bar{e}_g$ is equal to $\bar{e}_{g^{-1}}$; in particular every element of $\Gamma$ is a product of some of the $\bar{e}_g$ (without inverses). Since in $\Gamma$ we have $\bar{e}_g\bar{e}_h=\bar{e}_{gh}$, this shows that every element of $\Gamma$ is equal to $\bar{e}_g$ for some $g\in G$ and $\bar{e}_1=1$, and it immediately follows that $\mathrm{Ker}(\phi)$ is trivial. Hence $\phi$ is an isomorphism. $\Box$

Thus, the normal subgroup generated by $R$ is equal to the kernel $K$ of the canonical homomorphism $F_G\to G$ (mapping $e_g\mapsto g$).

The other (independent) step is to show that the subgroup $L$ of $F_G$ generated by $R$ is normal in $F_G$:

Indeed, we have the formula $$e_zf(x,y)e_z^{-1}=f(z,x)f(zx,y)f(z,xy)^{-1},$$ showing that $e_zLe_z^{-1}\subset L$.

Now let $U$ be the set of $w\in F_G$ such that $wLw^{-1}\subset L$. To show that $L$ is normal, we have to show that $U=F_G$. We know that $U$ is a subsemigroup of $F_G$, contains the positive generators, and contains $L$. It remains to show that it contains the inverses of generators. Indeed, we have $$e_z^{-1}=e_z^{-1}e_{z^{-1}}^{-1}e_{z^{-1}}=f(1,1)^{-1}f(z^{-1},z)^{-1}e_{z^{-1}}\in U.\quad\Box$$

We can now combine both steps: since $L$ is normal (by the second step) and contains $R$, it contains the normal subgroup generated by $R$, which by the first step equals $K=\mathrm{Ker}(\mathrm{can})$. So $L=K$.

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  • $\begingroup$ .@YCor I delete my answer in favor of yours, which is more complete. Thank you for the interaction. $\endgroup$ Apr 19, 2015 at 11:20
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    $\begingroup$ "and this is an isomorphism" -- why? $\endgroup$ Apr 19, 2015 at 14:04
  • $\begingroup$ .@darijgrinberg I think YCor shoud have written $\tilde{e_g}\mapsto g$ because $\tilde{e_g}$ is taken in the quotient of the free group. Now, we have a (set-theoretical) maps $g\mapsto e_g\mapsto \tilde{e_g}$ and the composite $g\mapsto \tilde{e_g}$ is easily shown to be the reciprocal of the first arrow. $\endgroup$ Apr 19, 2015 at 16:38
  • $\begingroup$ @DuchampGérardH.E. thanks too. You're right I used $e_g$ both in $F_G$ and in the quotient. $\endgroup$
    – YCor
    Apr 19, 2015 at 20:45
  • $\begingroup$ @YCor - thank you very much for your nice and detalied proof. It is the key point of proving the (I should say trivial :) ) theorem of Schur concerning the existence of the universal central extension of a perfect group. I have included detalis on other comment above. $\endgroup$ Apr 20, 2015 at 7:07
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Decided to add a separate answer for something mentioned in comments: the positive answer follows (almost) immediately from the Nielsen-Schreier theorem. I think it is worth it also because what one gets is a system of free generators for $\mathrm{Ker(can)}$: what that theorem actually immediately gives is that $\mathrm{Ker(can)}$ is freely generated by $e_1$, $e_xe_{x^{-1}}$ and $e_xe_ye_{xy}^{-1}$ for $x\in G\setminus\{1\}$, $y\in G\setminus\{x^{-1}\}$. This then obviously implies what we need since $e_1=e_1e_ye_{1y}^{-1}$ and $e_xe_{x^{-1}}=(e_xe_{x^{-1}}e_{xx^{-1}}^{-1})e_1$.

The theorem in full generality states that for a group $F$ freely generated by a set $X$, any subgroup $K$ of $F$ is freely generated by the set $$ \left\{\tau(Kw)x\tau(Kwx)^{-1}\mid w\in F,x\in X\right\}\setminus\{1\} $$ where $\tau:K\backslash F\to F$ is any section of the quotient map $F\twoheadrightarrow K\backslash F$ which gives a Schreier transversal, i. e. has the property that the image of $\tau$ is closed under taking left prefixes.

To obtain what we want just take $X=G$, $F=F_G$, $K=\mathrm{Ker(can)}$ and let $\tau: G\to F_G$ be given by $\tau(x)=e_x$ for $x\ne1$ and $\tau(1)=1$.

For an elementary proof of the Nielsen-Schreier see e. g. "An elementary proof that subgroups of free groups are free" by Benjamin Steinberg.

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Yes $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

Later

After quite some effort several people managed to convince me in the comments that the question is not as trivial as my answer seemed to imply. So I felt obliged to acknowledge it.

Still let me try to give an argument which hopefully shows why I was inclined to give that particular answer.

The subgroup $K$ generated by the words $e_xe_ye_{xy}^{-1}$ is clearly in the kernel.

To show it coincides with the kernel it suffices to show that the resulting surjective map $K\backslash F_G\twoheadrightarrow G$ (from left cosets to $G$) is injective.

To show that it suffices to reduce any word from $F_G$ to a single letter using only left multiplications by elements of $K$.

For all $x,y\in G$ the subgroup $K$ contains $e_{xy}\left(e_xe_y\right)^{-1}=\left(e_xe_ye_{xy}^{-1}\right)^{-1}$;

it also contains $e_1=e_1e_1e_{1\cdot1}^{-1}$;

hence it also contains $e_{x^{-1}}e_x=\left(e_{x^{-1}}e_xe_{x^{-1}x}^{-1}\right)e_1$.

Using this, given any word in the $e_x$es, start reading it from the left and do the following:

encountering an $\left(e_x\right)^{-1}$, multiply from the left by $e_{x^{-1}}e_x$ and cancel - as a result, the leftmost $\left(e_x\right)^{-1}$ will become replaced with $e_{x^{-1}}$;

encountering an $\left(e_1\right)^{\pm1}$, multiply from the left by $\left(e_1\right)^{\mp1}$ - as a result, the leftmost $\left(e_1\right)^{\pm1}$ drops out;

encountering $e_xe_y$, multiply from the left by $e_{xy}\left(e_xe_y\right)^{-1}$ - as a result, the leftmost $e_xe_y$ becomes replaced with $e_{xy}$.

Clearly one ends up either with a single $e_x$ or with the empty word which may (if one wishes) be replaced with $e_1$.

Now let me ask - is not all this ridiculously superfluous? :D

Still slightly later

Oops it remains to explain what to do with leftmost $e_x\left(e_y\right)^{-1}$ - in that case one has to multiply from the left by $e_{xy^{-1}}e_y\left(e_x\right)^{-1}=e_{xy^{-1}}e_y\left(e_{xy^{-1}\cdot y}\right)^{-1}$.

So I actually have to admit that I cannot make it any simpler than YCor's answer.

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    $\begingroup$ @HJRW Depends on whether users feel the answer was useful. Usually an answer of "yes" with no explanation is not considered useful. $\endgroup$
    – Todd Trimble
    Apr 18, 2015 at 17:04
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    $\begingroup$ @ToddTrimble, I think this answer is more useful than the question, which should probably have been closed as 'not research level'. $\endgroup$
    – HJRW
    Apr 18, 2015 at 22:01
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    $\begingroup$ @HJRW Well, if that's the case, then the question would have been better left unanswered. Answering a question can be seen as tacit approval of the question as appropriate for this site. We should not be sending mixed messages. (That is to say: if a question is to be answered at all, it should be answered seriously, not flippantly.) $\endgroup$
    – Todd Trimble
    Apr 18, 2015 at 22:27
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    $\begingroup$ PS: On the oher hand your opinion is opposite to the one of Ycor :) : ''The question is thus clearly non-trivial and should not be closed''. These facts puzzled me. $\endgroup$ Apr 19, 2015 at 6:26
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    $\begingroup$ And having written down my argument I also admit it is not entirely trivial :D $\endgroup$ Apr 20, 2015 at 7:36

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