# 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.

Ben Lynn blynn@cs.stanford.edu 💡