We need to be worried about whether this operation is independent of choice. That is, if \([x] = [x']\) and \([y] = [y']\) then must \([xy] = [x'y']\text{?}\) In other words, if \(x\sim x'\) and \(y \sim y'\text{,}\) must \(xy \sim x'y'\text{?}\)
As it turns out (serendipidous, I know) we discovered the exact property a subgroup needs to ensure its cosets form a well-defined group. At least, normally they do... 1
Get it?
Theorem3.30.Quotients and Normal Subgroups.
Let \(N\leq G\text{.}\) The set of (left cosets) \(G/N\) forms a group if and only if \(N\nsg G\text{.}\)
Definition3.31.Quotient Group.
For any normal subgroup \(N\) of a group \(G\text{,}\) the quotient group, \(G/N\text{,}\) is the set of left cosets of \(N\) in \(G\) with multiplication given by \(xN\cdot yN=(xy)N\text{.}\)
Example3.32.Old Friends.
Let \(H = \langle n \rangle\) in \(G = (\mathbb{Z}, +)\text{.}\) It is automatic that \(H\) is normal since \(G\) is abelian. The quotient group \(G/H\) is our old friend \((\mathbb{Z}/n, +)\text{.}\)
Remark3.33.
Don’t forget that the set \(G/H\) exists for any subgroup \(H\leq G\text{.}\) However, this set only satisfies the axioms of a group when the subgroup is normal.
Discussion3.2.Modular Integer? Barely Know ’Er!
Argue amongst yourselves what the ’best’ notation for the integers \(\mod n\) should be: \(\Z/n\)\(\Z/n\Z\text{,}\)\(\Z/n\text{,}\)\(\Z_n\text{,}\)\(\Z/\igen n\text{,}\)\(\frac{\Z}{n}\text{,}\) or whatever other horrors you can conjure.
Example3.34.Dihedral Quotient.
\(H = \langle r \rangle\) is a normal subgroup of \(D_{2n}\text{.}\) The quotient \(D_{2n}/H\) has just two elements, \(H\) and \(sH\text{.}\)
Exercise3.35.Quotient of Commutator Subgroup.
Let \(G\) be a group with commutator subgroup \(G'\text{.}\) Prove \(G/G'\) is abelian.
Exercise3.36.Special Linear Quotients.
\(\GL_n(\R)/\SL_n(\R)\cong\R^\times\)
Solution.
Define a map \(f: F \setminus \{0\}\to G/H\) as the composition of
\begin{equation*}
F \setminus \{0\} \xrightarrow{g} \operatorname{GL}_n(F) \twoheadrightarrow\operatorname{GL}_n(F)/\operatorname{SL}_n(F)
\end{equation*}
where \(g(x)\) is the \(n \times n\) matrix with \(x\) in the upper-left corner, \(1\)’s along the rest of the diaganal, and \(0\)’s everywhere else, and the second map is the canonical surjection. Then \(g\) is easily seen to be a homomorphism and hence so is \(f\) (since the composition of homomorhisms is a homomorphism). I claim \(f\) is one-to-one and onto. For any \(A \in \operatorname{GL}_n(F)\text{,}\) let \(x = \det(A)\) and note that \(g(x)^{-1} \cdot A\) has determinant \(1\) and thus belongs to \(\operatorname{SL}_n(F)\text{.}\) So \(f(x) = g(x)\operatorname{SL}_n(F) = A \operatorname{SL}_n(F)\text{.}\) This proves \(f\) is onto. If \(f(x) = e\text{,}\) then \(g(x) \operatorname{SL}_n(F) = \operatorname{SL}_n(F)\) and so \(g(x) \in \operatorname{SL}_n(F)\text{.}\) This means \(x = 1\text{.}\)
If \(G\) is finite we have \(\left | G/N\right |=\frac{|G|}{|N|}\text{.}\)
Definition3.38.Quotient Map.
For any group \(G\) and normal subgroup \(N\) of \(G\) the quotient map\(q:G \to G/N\) is defined by \(q(g)=gN\text{.}\)
Proposition3.39.Quotient Map is Surjective Homomorphism.
For any group \(G\) and normal subgroup \(N\) of \(G\text{,}\) the map \(q:G \to G/N\) defined by \(q(g)=gN\) is a surjective group homomorphism with kernel \(\operatorname{ker}(q)=N\text{.}\)
Proof.
Surjectivity is immediate from the definition. The group homomorphism property follows from the computation below which uses the definition of \(\pi\) and the rule for multiplying cosets in \(G/N\text{:}\)
Finally, using Lemma 3.4, we have \(\operatorname{Ker}(\pi)=\{g\in G\mid gN=e_GN\}=N\text{.}\)
Theorem3.40.Normal Subgroup iff Kernel.
A subgroup \(N\) of a group \(G\) is normal in \(G\) if and only if \(N\) is the kernel of a homomorphism with domain \(G\text{.}\)
Proof.
First, suppose \(K\nsg G\text{.}\) Let \(H=G/K\) (which is a group since \(K\) is normal in \(G\)) and define \(\varphi: G\to G/K\) such that \(\varphi(g)=gK\text{.}\) Let \(k\in K\text{,}\) and observe that \(\varphi(k)=kK=K\text{,}\) and so \(k\in\ker(\varphi)\text{.}\) Let \(g\in\ker(\varphi)\text{,}\) meaning that \(\varphi(g)=K\text{,}\) placing \(g\in K\text{.}\) By Proposition 3.39 this map is a homomorphism.
Next, suppose there exists a group \(H\) and and a homomorphism \(\varphi : G\to H\) such that \(K = \ker(\varphi)\text{.}\) Let \(g\in G\) and consider \(gKg\inv\text{.}\) Observe
given that \(\ker(\varphi)=K\text{.}\) Thus \(gKg\inv\subseteq K\) for all \(g\in G\text{,}\) making \(K\) a normal subgroup of \(G\text{.}\)
Qual Watch.
Proving Theorem 3.40 was Part (a) of [cross-reference to target(s) "jan-2022-3" missing or not unique] on the [cross-reference to target(s) "jan-2022" missing or not unique] qualifying exam.
Theorem3.41.Cyclic Center Quotients.
Let \(G\) be a group with center \(Z(G)\text{.}\) If \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Summary
Given \(N\nsg G\text{,}\) the Quotient Group\(G/N\) is the set of left cosets of \(N\) in \(G\) with multiplication given by \(xN\cdot yN=(xy)N\text{.}\) This is a well defined group if and only if \(N\) is normal. 2
