Group Theory

Suppose we have a set of symbols $\{x_1,...,x_n\}$. Consider the words we may form from them, that is, formal products of the form $x_{i_1}^{a_1}...x_{i_k}^{a_k}$ where the exponents are integers (and may be negative). The empty word is denoted by 1. A word is reduced if it is empty or no two consecutive $x$'s have the same subscript. Define multiplication on words by concatenation. We may reduce a word by using the rules $x^a{x}^b=x^{a+b}$ and $x^0=1$.

It can be shown that we have constructed a free group in this manner. The only nontrivial fact to verify is that concatenation is indeed associative, which is tedious and will be omitted.

Now consider a group $G$ that is generated by $n$ elements $g_1,...,g_n$. Then consider the map from the free group $F$ generated by $n$ elements that sends $x_i$ to $g_i$. The kernel of this map $R$ consists precisely of nontrivial relations $r(x_1,...,x_n)$ such that $r(g_1,...,g_n)=1$. Summarizing:

Theorem: Every group $G$ which can be generated by $n$ elements can be represented as the homomorphic image of the free group $F$ on $n$ generators. The kernel of this map consists of elements of $F$ that correspond to relations in $G$.

The groups $F,R$ are said to form a presentation of $G$. Conversely given any normal subgroup $R$ of a free group $F$, we may form a group $F/R$.

Now suppose we are given $m$ relations $r_1,...,r_m$ on $n$ elements $x_1,...,x_n$. The group consisting of the smallest normal subgroup of $F$ that contain all $m$ relations is denoted by

$$R=\{r_1,...,r_m{\}}^F$$

and is called the normal closure of $r_1,...,r_m$ and may be called the relation group of $G=F/R$.

Now suppose $H$ is another group with generators $g_1,...,g_n$ that satisfies all the relations that $G$ does, but in addition also satisfies relations $t_1,...,t_p$. Then consider $S=\{r_1,...,r_m,t_1,...,t_p{\}}^F$. We have $H=F/S$. Since $S\supset R$ as in the third isomorphism theorem we may view $A=S/R$ as a normal subgroup of $F/R$, and we have $H\cong G/A$, thus:

Theorem: If new relations are added to a group $G$, the resulting group is a homomorphic image of $G$.

Hence $F/R$ is the freest group with $n$ generators satisfying given relations $r_1,...,r_m$.

As an application, we can make a group $G$ abelian by considering $G/G\prime$ where $G\prime$ is the normal closure of relations of the form $g_i^{-1}{g}_j^{-1}{g}_i{g}_j$ for all $i,j$.

Example: Let $G$ be the quarternion group $⟨a,b\mid a^4=1,a^2=b^2,ba=a^{3}b⟩$. Then $G/G\prime$ is generated by $u=aG\prime ,v=bG\prime$. In additive notation we have $4u=0,2u=2v,v+u=3u+v$, thus $2u=2v=0$ and we find $G/G\prime ={\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

If the free group on $x_1,...,x_n$ is made abelian then we obtain the free abelian group on $x_1,...,x_n$. This implies that free groups on different numbers of generators cannot be isomorphic, otherwise we would have their abelian counterparts isomorphic, a contradiction by a previous result.

Fact: A subgroup of a free group that contains more than one element is a free group.