I'm working through Aluffi's Algrebra: Chapter 0 and I need some assistance with an excercise.

Aluffi, Ex. II.8.7

Let $(A|R)$, resp. $(A'|R')$, be a presentation for a group $G$ in Grp, resp. $G'$; we may assume that $A, A'$ are disjoint. Prove that the group $G \ast G'$ presented by \begin{equation} (A \cup A'|R \cup R') \end{equation} satisfies the universal property for the coproduct of $G$ and $G'$ in Grp. (Use the universal properties of both free groups and quotients to construct natural homomorphisms $G \to G \ast G', G' \to G \ast G'.$)

What I have done so far:

First off, we have natural set-functions $A,A' \to F(A \cup A')$. Hence we have unique homomorphisms $F(A),F(A') \to F(A \cup A')$ whose restrictions to $A,A'$ are equal to the functions $A,A' \to F(A \cup A')$, by the universal property of free groups.

By combining these homomorphisms with the natural projection $F(A \cup A') \to F(A \cup A')/(R \cup R')$, we get homomorphisms $F(A),F(A') \to F(A \cup A')/(R \cup R')$.

Since $a(R) = b(R) \Rightarrow a(R \cup R') = b(R \cup R')$, we get a natural homomorphism $\pi_1:G \cong F(A)/(R) \to F(A \cup A')/(R \cup R')$, by the universal property of quotients. Similarly, we get a natural homomorphism $\pi_2:G' \cong F(A')/(R') \to F(A \cup A')/(R \cup R') = G \ast G'$.

If I'm not mistaken, these homomorphisms simply map $a(R) \mapsto a(R \cup R')$, for all $a(R) \in G$, and similarly for elements of $G'$. If we have a group $H$ and homomorphisms $f:G \to H, g:G' \to H$, then the only function $\varphi:G \ast G' \to H$ s.t. $\varphi \circ \pi_1 = f, \varphi \circ \pi_2 = g$, must map $a(R \cup R') \mapsto f(a(R))$ for all $a \in F(A)$, and similarly for all $b \in F(A'), b(R \cup R') \mapsto g(b(R'))$. All that is left, is to show that this defines a function $\psi$.

So, we need to show that $ab^{-1} \in (R \cup R') \Rightarrow \psi (ab^{-1}(R \cup R')) = e_H$, for any $a,b \in F(A \cup A')$.

This is what I can't do. How do I go about showing this, or am I going in a wrong direction?

Edit: Note that (R) stands for the normal closure of R in F(A).

One answer:

Note that $(R \cup R') := \langle \{ghg^{-1} \mid g \in F(A \cup A'), h \in R \cup R'\} \rangle$. Now, for any $h \in R \cup R', g \in F(A \cup A')$, we have \begin{align} \psi (ghg^{-1}(R \cup R')) & = \psi (g(R \cup R')) \psi (h(R \cup R')) \psi (g^{-1}(R \cup R')) \\ & = \psi (g(R \cup R')) \psi (g^{-1}(R \cup R')) \\ & = \psi (gg^{-1}(R \cup R')) \\ & = e_H. \end{align} Hence, for any $a \in (R \cup R')$, we have $\psi (a) = e_H$, which completes the proof.


I guess by $(R)$ you mean the normal closure $\langle R^{F(A)} \rangle$ of $R$ in $F(A)$?

There is a unique homomorphism $\psi:F(A \cup A') \to H$ with $\psi(a) = f(a)$ for $a \in A$ and $\psi(a) = g(a)$ for $a \in A'$. Since all elements of $R$ lie in $\ker(f)$ and all elements of $R'$ lie in $\ker(g)$, we have $R \cup R' \subset \ker(\psi)$ and hence, since $\ker(\psi) \unlhd F(A \cup A')$, we have $(R \cup R') \le \ker(\psi)$.

So $\psi$ induces a well-defined map $\varphi:G*H = F(A \cup A')/(R \cup R') \to H$, and $\varphi$ satisfies $\varphi \circ \pi_1 = f$ and $\varphi \circ \pi_2 = g$, as required.

  • $\begingroup$ Yes, I mean the normal closure. What about the case where $a \in A, b \in A'$? We could still have $ab^{-1} \in (R \cup R')$. $\endgroup$ – VJP May 27 '15 at 15:50
  • $\begingroup$ No you couldn't because the element sof $R$ are words over $A \cup A^{-1}$, and the elements of $R'$ are words over $A' \cup A'^{-1}$, and you assumed that $A$ and $A'$ are disjoint. $\endgroup$ – Derek Holt May 27 '15 at 16:33
  • $\begingroup$ What about $a \in R, b \in R'$? We would still have $ab^{-1} \in (R \cup R')$. And in any case, how would I extend the condition from all $a,b \in A \cup A'$ to all $a,b \in F(A \cup A')$? $\endgroup$ – VJP May 27 '15 at 17:03
  • $\begingroup$ I accepted this as an answer, but I also wrote down an answer I came up with. Thank you for your help Derek. $\endgroup$ – VJP May 27 '15 at 17:30
  • $\begingroup$ In your answer, your definition of $(R \cup R')$ is wrong. It is not the set of conjugates of elements of $R \cup R'$, but the subgroup generated by that set. $\endgroup$ – Derek Holt May 27 '15 at 20:38

Often it is much more convenient to work with representable functors.

The universal property of the group $(A|R)$ is the following: If $H$ is a group, there is a natural isomorphism (i.e. bijection) $$\hom((A|R),H) \cong \{f \in \hom(A,|H|) : f(R)=\{1\}\},$$ where by $f(R)$ I mean the set of all $f(r)$, $r \in R$, and $f(r)$ is defined by extension of $f$ on the set of words. It follows for test groups $H$: $$\hom((A \sqcup A'|R \sqcup R'),H)$$ $$ \cong \{g \in \hom(A \sqcup A',|H|) : g(R \sqcup R')=\{1\}\}$$ $$= \{g \in \hom(A \sqcup A',|H|) : g(R)=\{1\} \wedge g(R')=\{1\}\}$$ $$\cong \{(f,f') \in \hom(A,|H|) \times \hom(A',|H|) : f(R)=\{1\} \wedge f'(R')=\{1\}\}$$ $$ \cong \hom((A|R),H) \times \hom((A'|R'),H).$$ These isomorphisms are clearly natural in $H$. Thus, by definition of a coproduct, $(A \sqcup A'|R \sqcup R')$ is a coproduct of $(A|R)$ and $(A'|R')$.


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