\beginsection Free Groups

It might be nice  to have aniversal group.  

Let $S$ be a set. By the {\bf free group} generated by $S$
I mean the set of all expressions
$$s_1^{n_1}s_2^{n_2}\cdots s_m^{n_m},\eqno (*)$$
where $s_i\in S$ and $n_i\in\Z$,
subject to the identifications\sk
$$s_1^{n_1}\cdots s_k^{n_k}s_{k+1}^{0}s_{k+2}^{n_{k+2}}\cdots s_m^{n_m}=s_1^{n_1}\cdots s_k^{n_k}s_{k+2}^{n_{k+2}}\cdots s_m^{n_m}$$
$$s_1^{n_1}\cdots s_k^{n_k}s_{k+1}^{k+1}\cdots s_m^{n_m}=s_1^{n_1}\cdots s_k^{n_k+n_{k+1}}\cdots s_m^{n_m}$$
if $s_k=s_{k+1}$.  Call the empty product $1$.     We define an operation this set by setting 
$$(s_1^{n_1}s_2^{n_2}\cdots s_m^{n_m})(t_1^{p_1}t_2^{p_2}\cdots t_k^{p_k})=
s_1^{n_1}s_2^{n_2}\cdots s_m^{n_m}t_1^{p_1}t_2^{p_2}\cdots t_k^{p_k}.$$
Call this set with an operaton $F[S]$.\sk After making the above identifiations, every element of $F(S)$ has a unique representation in the form $(*)$ such that $n_i\ne0$ and $s_j\ne s_{j+1}$.

\proclaim  Theorem.  $F[S]$ is a group and if $G$ is a group generated by at most $|S|$ elements there exists a surjective homomorphism from $F[S]$ onto $G$.  In fact, if $f\colon S\ra G$ is a function such that $\dia{\stt{f(s)\colon s\in S}}=G$ there exists a unique surjective homomorphism $\phi_f\colon F(S)\ra G$ such that 
$$\phi_f(s^1)=f(s)\quad \hb{for } s\in  S$$

\sk\pr \vsk1

Now set $\phi_f(s_1^{n_1}s_2^{n_2}\cdots s_m^{n_m})=f(s_1)^{n_1}f(s_2)^{n_2}\cdots f(s_m)^{n_m}.$\vsk1


Now let $C(S)$ be the smallest subgroup of $F(S)$ containing $aba\iv b\iv$ for all $a,b\in F(S)$.

\proclaim Lemma.  $C(S)$ is normal and the factor group $A(S)$ is the ``largest'' Abelian quotient of $F(S)$. 

\pr\vsk2


The group $A(S)$ is called the {\bf free Abelian group generated by $S$} and any g      ou isomorphic to one of these is called {\bf  free Abelian}.  

\proclaim Theorem.  If $G$ is an Abelian group generated by at most $|S|$ elements there exists a surjective homomorphism from $A(S)$ onto $G$.

\pr\vsk1

\proclaim Corollary. Suppose $S$ is finite. Then $A(S)\isom \Z^{|S|}$.

\pr\sk




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