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Suppose that $G=\langle S\mid R\rangle$ (where $S\subset G$ is a subset with a set of relations $R$) and that $\psi$ is a function $S\to H$ where $H$ is an arbirtrary group. Suppose further that each relation in $R$ becomes $1\in H$ if each $T\in S$ that appears in the relation is replaced by $\psi(T)$.

Show that $\psi$ can be extended to a homomorphism $G\to H$.

I have trouble understanding the concepts of relations. It is clear that the proof somehow follows from the universal property of free groups. But how exactly so? I know that $T_1^{\epsilon_1}\cdots T_k^{\epsilon_k}\in R$ maps to $\psi(T_1)^{\epsilon_1}\cdots \psi(T_k)^{\epsilon_k}=1\in H$, $\epsilon_i\in\{\pm 1\}$ but what can I derive from that other than information about the kernel as soon as I know it's a homomorphism?

I've also found this paragraph in my book:

If $G$ is an arbitrary group generated by a subset $S$, then there is a homomorphism $\phi$ from a free group $F$ of rank $|S|$ onto $G$, and so $G\cong F/\ker(\phi)$. In fact, $F$ may be chosen to be the free group based on the set $S$ itself. If an element $T\in S$ is denoted by $\hat T$ when it is considered as an element of the free group $F$ rather than as an element of $G$, then the homomorphism is the map that results from setting $\phi(\hat T)=T$ for all $\hat T\in S$.

I'm fairly sure everything I need can be found in there but how do I glue it together? Since $H$ was arbirtrary I can't treat it as the free group based on $S$.

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user424862
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  • The ingredients are to write $G$ as a quotient of a free group $F$, use $\psi$ to define a homomorphism from $F$ to $H$, and use the universal property of quotients. – Elle Najt Jul 02 '17 at 09:28
  • Why don't you try and define the homomorphism first? Given $g \in G$, we can write it as a product of the generators i.e. $g = s_1^{a_1} s_2^{a_2}...s_n^{a_n}$. Now define the homomorphism by $\psi(g) = \psi(s_1)^{a_1}...\psi(s_n)^{a_n}$. Note that this is well defined because it's defined in a way that respects the operation, and respects relations. – Osama Ghani Jul 02 '17 at 09:28

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The definition of $\langle S\mid R\rangle$ is supposedly that $\langle S\mid R\rangle=F/N$ where $F$ is a free group over $S$ and $N$ is the intersection of all normal subgroups of $F$ containing $R$.

The definition of $F$ (strictly speaking, together with a map $i\colon S\to F$) is by a universal property: For every group $H$ and every map $f\colon S\to H$, there exists one and only one homomorphism $\phi\colon F\to H$ such that $\phi\circ i=f$.

The definition of $F/N$ (strictly speaking, together with a projection $\pi\colon F\to F/N$) is also via a universal property: For every group $H$ and homomorphism $\phi\colon F\to H$ with $\phi\restriction_N=0$, there exists one and only one homomorphism $\bar\phi\colon F/N\to H$ such that $\bar\phi\circ \pi=\phi$.

Now in your problem, we are given a map $\psi\colon S\to H$. As seen above, this extends in a unique fashion to a homomorphism $F\to H$. As "replacing every $T\in S$ with $\psi(T)$" defines a homomorphism $F\to H$, this must be said unique homomorphism (and we call it $\psi$ as well). By the given condition, $R\subseteq \ker\psi$, hence $N\le \ker\psi$. It follows that we obtain a unique homomorphism (again recycling the name) $\psi\colon \langle S\mid R\rangle \to H$ as desired.

Hagen von Eitzen
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  • Thanks for explaining so well! I now understand that we have a unique homomorphism $\psi: F\to H$ that we want to use to construct a homomorphism $\psi': G \to H$.From the quoted paragraph we also know that $F/\ker(\psi)\cong H$. So in conclusion we can project $\pi\circ\psi: F\to F/\ker(\psi)\to H$, right?

    But since $F/\ker/(\psi)\neq G$ we have to do something more and I don't understand how $\langle S\mid R\rangle =F/N$ with your specific $N$.

     – user424862 Jul 02 '17 at 10:44
  • Reading the thread (link) I now understand that $G=F/N$. So now the only thing left I can't comprehend yet are your last two sentences. – user424862 Jul 02 '17 at 12:09