## Group Presentations

We defined a lot of terms on the Words Over a Set page. Recall that if $A$ is a nonempty set and $\mathcal R$ is a set of words over $A \cup A^{-}$ then we defined an equivalence relation $\sim$ on the set of words over $A \cup A^{-}$ for words $u$ and $v$ as follows. We say that $u \sim v$ if there exists a finite sequence of words $w_1, w_2, ..., w_n$ with $w_1 = u$ and $w_n = v$ such that for each $i \in \{ 1, 2, ..., n - 1\}$:

**1)**$w_{i+1}$ is obtained from $w_i$ by inserting an occurrence of $xx^{-1}$ in the word $w_i$.

**2)**$w_{i+1}$ is obtained from $w_i$ by deleting an occurrence $xx^{-1}$ in the word $w_i$.

**3)**$w_{i+1}$ is obtained from $w_i$ by inserting a word in $\mathcal R$ or $\mathcal R^{-}$ in the word $w_i$.

**4)**$w_{i+1}$ is obtained from $w_i$ by deleting a word in $\mathcal R$ or $\mathcal R^{-}$ in the word $w_i$.

Let $\langle A, \mathcal R \rangle$ denote the set of equivalence classes of words over $A \cup A^{-}$. Define an operation on these equivalence classes for all $[u], [v] \in \langle A, \mathcal R \rangle$ by:

(1)It can be shown that this operation is well-defined. Furthermore, for all equivalence classes $[u], [v], [w] \in \langle A, \mathcal R \rangle$ we have that:

**1)**$[u]([v][w]) = ([u][v])[w]$. (The Associativity Property)

**2)**$[u][()] = [u]$ and $[()][u] = [u]$ where $[()]$ is the equivalence class of the empty word $()$. (The Existence of an Identity)

**3)**$[u][u^{-1}] = [()]$ and $[u^{-1}[u] = [()]$. (The Existence of Inverses)

Therefore $\langle A, \mathcal R \rangle$ with this operation forms a group.

Definition: Let $A$ be a nonempty set and let $\mathcal R$ be a set of words over $A \cup A^{-}$. Let $G$ be a group. Let $f : A \to G$ be any function. We define a similar function $f : \{ \mathrm{words \:over \:} A \cup A^{-} \} \to G$ for all $u = a_1^{\epsilon_1}a_2^{\epsilon_2}...a_n^{\epsilon_n}$ where $a_1, a_2, ..., a_n \in A$ and $\epsilon_1, \epsilon_2, ..., \epsilon_n \in \{ +, - \}$ by:$f(u) = f(a_1^{\epsilon_1}a_2^{\epsilon_2}...a_n^{\epsilon_n}) := f(a_1)^{\epsilon_1}f(a_2)^{\epsilon_2}...f(a_n)^{\epsilon_n}$. And further define another similar function $f : \langle A, \mathcal R \rangle \to G$ by: $f([u]) = f([a_1^{\epsilon_1}a_2^{\epsilon_2}...a_n^{\epsilon_n}]) = f([a_1])^{\epsilon_1}f([a_2])^{\epsilon_2}...f([a_n])^{\epsilon_n}$. |

If the function $f : \langle A, \mathcal R \rangle \to G$ is well-defined then $f$ is a group homomorphism. However, this is not always the case. The following theorem gives us a sufficient and necessary condition for determining when $f$ is well-defined.

Theorem 1: Let $A$ be a set and let $\mathcal R$ be a set of words over $A \cup A^{-}$. Then the function $f : \langle A, \mathcal R \rangle \to G$ above is well-defined if and only if for every word $R \in \mathcal R$ we have that $f([R]) = 1$ where $1$ is the identity in $G$. |

**Proof:**$\Rightarrow$ If $f$ is well-defined then $f$ is a group homomorphism. Then for all $R \in \mathcal R$ we have that $[R] = [()]$. But group homomorphisms send the identity of one group to the identity of the other group. So indeed, $f([R]) = 1$.

- $\Leftarrow$ Let $[u], [v] \in \langle A, \mathcal R \rangle$ and that $[u] = [v]$. Then $f([u]) = f([v])$ if $v$ can be obtained from $u$ from either of the first two steps mentioned at the top of this page. For the steps $3$ and $4$, if $u = w_1w_2$ and $v = w_1Rw_2$ then:

- So indeed, $f$ is well-defined regardless of the representative chosen for the equivalence class $[u]$. $\blacksquare$

Definition: Let $A$ be a nonempty set and let $\mathcal R$ be a set of words over $A \cup A^{-}$. Let $G$ be a group. If $f : \langle A, \mathcal R \rangle \to G$ is well-defined (and hence a homomorphism) and is further a group isomorphism, then $\langle A, \mathcal R \rangle$ is said to be a Group Presentation of $G$. If $A$ and $\mathcal R$ are both finite sets then this group is said to be Finitely Presented. |