“Megamind: Oh, you’re a villain, alright, just not a super one.” “Tighten: Yeah? What’s the difference?”
“Megamind: Presentation!”
―Megamind (2010)
Definition4.4.Generated Normal Subgroup.
Let \(G\) be a group and let \(R \subseteq G\) be a set. The normal subgroup of \(G\) generated by \(R\), denoted \(\langle R \rangle ^N\text{,}\) is the set of all products of conjugates of elements of \(R\) and inverses of elements of \(R\text{.}\) In symbols,
Let \(A\) be a set and let \(R\) be a subset of the free group \(F(A)\text{.}\) The group with presentation\(\langle A \mid R \rangle = \langle A | \{r = e \mid r \in R\} \rangle\) is defined to be the quotient group \(F(A)/\langle R \rangle ^N\text{.}\)
Example4.6.
For \(A=\{x\}\text{,}\)\(R=\{x^n\}\) we obtain the cyclic group of order \(n\text{:}\)
Let \(A\) be a set, let \(F(A)\) be the free group on \(A\text{,}\) let \(R\) be a subset of \(F(A)\text{,}\) let \(H\) be a group, and let \(g:A \to H\) be a function satisfying the property that whenever \(r = a_1^{i1} \cdots a_m^{i_m} \in R\text{,}\) with each \(a_j \in A, g_j \in G\) and \(i_j \in \{1,-1\}\text{,}\) then \((g(a_1))^{i_1} \cdots (g(a_m))^{i_m} = e_H\text{.}\) Then there is a unique homomorphism \(\overline{f}: \langle A | R \rangle \to H\) satisfying \(\overline{f}(a\langle R\rangle ^N) = g(a)\) for all \(a \in A\text{.}\)
Proof.
By the UMP of the free group there is a unique group homomorphism \(\tilde f:F(A)\to H\) such that \(f(a)=g(a)\) for all \(a\in A\text{.}\) Then for \(r = a_1^{i1} \cdots a_m^{i_m} \in R\text{,}\) we have \(f(r)=(g(a_1))^{i_1} \cdots (g(a_m))^{i_m} = e_H\text{,}\) showing that \(R\subseteq \operatorname{Ker}(f)\text{.}\) Since \(\operatorname{Ker}(f)\mathrel{\unlhd}F(A)\) and \(\langle R\rangle ^N\) is the smallest normal subgroup containing \(R\text{,}\) it follows that \(\langle R\rangle ^N\subseteq \operatorname{Ker}(f)\text{.}\) By the UMP of the quotient, \(f\) induces a group homomorphism \(\overline{f}: G/\langle R\rangle ^N\to H\text{.}\) Moreover, for each \(a\in A\) we have \(g(a)=f(a)=\overline{f}(a\langle R\rangle ^N)\text{.}\)
Remark4.9.
The UMP of the presentation says that one can build a homomorphism from a group with a given presentation to any other group \(H\) as long as one is able to send the generators (elements of \(A\)) via some function \(g\) to some elements of \(H\) that satisfy the same relations in \(H\) as those given in the presentation.
Example4.10.
To find a groups homomorphism \(D_{2n}\to \operatorname{GL}_2({\mathbb{R}})\) it suffices to find a map \(g:\{r,s\}\to \operatorname{GL}_2({\mathbb{R}})\text{,}\) say \(r\mapsto R, s\mapsto S\) and to verify that \(S^2=I_2,R^n=I_2, SRSR=I_2\text{.}\) As you have shown on homework, this does hold for the matrices
By the UMP of the presentation there is a group homomorphism \(f:D_{2n}\to \operatorname{GL}_2({\mathbb{R}})\) that extends \(g\text{;}\) that is \(f(r^is)=R^iS\) for all \(0\leq i<n\text{.}\)
We can use the map above to understand the notions of isometry and orientation-preserving a little more rigorously. Indeed, an invertible linear transformation, equivalently an element \(M\in \operatorname{GL}_n({\mathbb{R}})\) is an isometry if and only if \(\det(M)=\pm 1\text{.}\) An invertible linear transformation, equivalently an element \(M\in \operatorname{GL}_n({\mathbb{R}})\) is orientation-preserving if and only if \(\det(M)>0\text{.}\) Thus the group of orientation-preserving isometries of \({\mathbb{R}}^n\) can be identified with \(\operatorname{SL}_n({\mathbb{R}})\text{.}\)
SubsectionEvery Group is a Quotient of a Free Group