## Generators and Relations

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 $G/G'$ where $G'$ 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 $\langle a,b | a^4 = 1, a^2 = b^2, b a = a^3 b \rangle$. Then $G/G'$ is generated by $u = a G' , v = b G'$. In additive notation we have $4 u = 0, 2u = 2v, v + u = 3u + v$, thus $2 u = 2 v = 0$ and we find $G/G' = \mathbb{Z}_2 \times \mathbb{Z}_2$.

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.