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Why is every group the quotient of a free group by a normal subgroup?

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I've added a question to this post. –  anon Nov 8 '10 at 15:07
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3 Answers

This is one of the most intuitive observations in all of group theory, and it illustrates the quotient operation in the most fundamental way.

I'll provide two separate answers. The first is fully intuitive; the second is a formalized version of the first.

First answer: Take a group $G$. A relation on $G$ is an equation satisfied by some of the elements. For instance, $eg = g$ where $e$ is the identity is a relation satisfied by all group elements $g \in G$. Because we can always multiply by inverses in a group, we can rewrite this relation as $egg^{-1} = gg^{-1} = e$, i.e., $e = e$. This can be applied to any relation. If $G$ is abelian, then $ab = ba$ for all $a,b \in G$, and we can rewrite this as $aba^{-1}b^{-1} = e$.

In other words, a relation asserts that some product of group elements coincides with the identity, so the only information we need to understand the relation is the product which occurs on the left side of the equals sign.

Now every group has a few relations which are implied directly by the group axioms. $aa^{-1} = e$ is one of them. We can ask whether the group has any extra relations which are not implied by the group axioms. If no such relations exist, i.e., if the only relations which hold are those which must hold by virtue of the group axioms, then the group is said to be free; the group is "free of additional relations."

If you have a group $G$, one natural thing to do is to introduce new relations into it and to thereby create new groups. But you can't just introduce completely random relations because (a) the relations can't contradict each other or pre-exising relations and (b) the resulting structure must again be a group. Now we saw earlier that a relation can be specified as a product of group elements. In order that the relations satisfy (a) and (b), it turns out it is necessary and sufficient that the corresponding products form a normal subgroup $N$. The result of introducing the collection of relations $N$ into the group $G$ is the quotient $G/N$.

Any group $G$ can be obtained in this manner. You start with the free group $F$ whose generators are elements of $G$ considered as a set. And then you look at all the additional relations satisfied by elements of $G$ and assemble them into a normal subgroup $N$. Then $G = F/N$ by the above.

Second answer: Given any set $S$, the free group on $S$ is that group $F(S)$ for which every function $f : S \rightarrow G$ from $S$ to an arbitrary group $G$ extends to a unique homomorphism $\tilde{f} : F(S) \rightarrow G$. There are various ways of constructing $F(S)$ explicitly. For instance, you may take $F(S)$ to consist of words over the alphabet whose letters are elements of $S$ and $S'$, where $S'$ has the letter $s^{-1}$ (a symbol at the moment) for each symbol $s \in S$. It's important to notice that $F(S)$ actually contains equivalence classes of words, because we introduce the obvious cancellation rules; e.g., $abb^{-1}c$ can be reduced via cancellation to $ac$. It must be proved that all possible algorithms for reduction yield the same reduced word; I'll omit that step.

You also have to prove that this group $F(S)$ satisfies the stated universal property. I won't prove this in detail, but it is more or less intuitive. Since $\tilde{f}$ has to be a homomorphism, we find, for instance, that $\tilde{f}(ab) = \tilde{f}(a) \tilde{f}(b) = f(a)f(b)$. In general, since $f$ is defined for all elements of $S$, $\tilde{f}$ is thereby defined uniquely for all elements of $F(S)$. [It is via similar reasoning that you may determine that it is sufficient to know the values of a linear operator on the elements of a basis of a vector space.]

So we start with our group $G$ which we would like to write as a quotient of a free group. Which free group? That free group whose generators are the symbols from $G$. So we pick $F(G)$. Now we need to introduce the needed relations in order to collapse $F(G)$ into $G$. How do we carry it out? By the first answer, we could easily accomplish this if only we knew the normal subgroup $N$ of relations, but it seems that in this general case we don't really know $N$ concretely.

In fact, we can figure out $N$ as follows. We can take the identity map $f : G \rightarrow G$ and extend it to a homomorphism $\tilde{f} : F(G) \rightarrow G$. The extension $\tilde{f}$ is in general not injective, and its kernel is precisely the group of relations $N$! (Formally this is an application of one of the standard theorems on homomorphisms.) Then $G = F(G)/N$ as before.

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In the last paragraph, you cannot in general 'extend' the set map from $G$ to $F(G)$ into a homomorphism (there is nothing to extend, and it will not be a homomorphism unless $G$ is trivial). Do you mean, perhaps, to map the set $G$ into the group $G$, and then extend that to a homomorphism $\overline{f}\colon F(G)\to G$? –  Arturo Magidin Nov 8 '10 at 19:03
    
Thanks very much for the correction! –  Zach Conn Nov 8 '10 at 19:10
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This was really, really helpful Zach. Thanks a ton. –  Pedro Tamaroff Feb 22 '13 at 22:59
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It is a consequence of the universal property of free groups.

If $X$ is a set, then the free group on $X$, $F(X)$, is a group $F(X)$ that contains $X$ as a subset, and such that for every group $G$ and every set-theoretic map $f\colon X\to G$ there exists a unique group homomorphism $\mathcal{F}\colon F(X)\to G$ such that $\mathcal{F}(x) = f(x)$ for all $x\in X$.

If $G$ is a group, let $X$ be the underlying set of $G$ (the set of all elements of $G$, considered merely as a set), and let $F(X)$ be the free group on $X$. The set-theoretic map $i\colon X\to G$ that maps each element of $G$ to itself induces a homomorphism $\mathcal{F}\colon F(X)\to G$ such that $\mathcal{F}(x)=i(x)$ for all $x\in X$. Since the image of $\mathcal{F}$ includes at least the image of $i$, the map $\mathcal{F}$ is onto. By the First Fundamental Theorem of Homomorphisms, the image of $\mathcal{F}$ (which is $G$) is isomorphic to $F(X)/\mathrm{ker}(\mathcal{F})$. So $G$ is isomorphic to the quotient of a free group by a normal subgroup.

You can replace the underlying set of $G$ with any generating set of $G$ and the proof goes through.

Added: Of course, this answer requires that you know that a free group exists on any set, and that it has the appropriate universal property. A good exposition, with motivation and three different constructions of the free group, is in Chapter 2 of George Bergman's book, An invitation to General Algebra and Universal Constructions. Here's a link to a Postscript version.

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Given any group G, let F(G) be the free group whose generators are the elements of G. Define a group homomorphism f from F(G) to G in the obvious way: by sending the generator g of F(G) to the element g of G. This is a surjective homomorphism, so G is isomorphic to F(G) / ker f.

In fact, let M be the set of words of the form $g_1 g_2 h^{-1}$ where $g_1 g_2 = h$ in G -- in other words, M encodes the multiplication table for G. Let N be the normal closure of M. Then G is isomorphic to F(G) / N.

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