# 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

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.

*blynn@cs.stanford.edu*💡