---- > [!definition] Definition. ([[quotient group]]) > Let $G$ be a [[group]] and $N$ a [[normal subgroup]]. We define the **quotient group** $G / N$ to be the collection of left (=right) [[coset|cosets]] of $N$ along with the [[binary operation]] $aN \star bN := (ab)N.$ > [!equivalence] Derivation. > ![[characterization of quotienting a group#^90a179]] ^equivalence > [!intuition] > ![[CleanShot 2023-09-11 at [email protected]]] > [!justification] > A more natural discussion in [[characterization of quotienting a group]]. But if in a hurry, we can also quickly just check that $G/N$ is indeed a [[group]] and move on... The identity element is $N$ $(=eN)$ itself. The inverse of $aN$ is $a^{-1}N$. One thing to be careful about is ensuring the group law is [[well-defined]]— that choice of [[coset]] representative does not matter. So suppose $aH=a'H, b'H=bH$, need to check that $abH=a'b'H$. Note that $a'=ah_{1}$ and $b'=bh_{2}$ for some $h_{1},h_{2} \in H$. Now $a'b'=ah_{1}bh_{2}=ab(b^{-1}h_{1}b)h_{2} \in abH$ > implying $a'b'H \subset ab H$ which suffices to complete the proof. > [!basicexample] > In the [[general linear group]] $GL_{3}(\mathbb{R})$, consider the subsets $H=\begin{bmatrix} 1 & * & * \\ & 1 & * \\ & & 1 \end{bmatrix}, K=\begin{bmatrix} 1 & 0 & * \\ & 1 & 0 \\ & & 1 \end{bmatrix}.$ $H$ is a [[subgroup]] of $G$, $K$ is a [[normal subgroup]] of $H$, and the [[quotient group]] $H / K$ is [[group isomorphism|isomorphic]] to $\mathbb{R}^{2}$. The [[center of a group|center]] of $H$ is $K$. Any [[matrix]] of $Hs form is [[inverse matrix|invertible]] by [[determining invertibility from upper-triangular matrix]] (all diagonal entries are nonzero), and hence belongs to $GL_{3}(\mathbb{R})$. In particular, since $\det H=1$, $H^{-1}$ [[matrix inverse is adjugate over determinant|equals the adjugate]] of $H$, which is [[upper-triangular matrix|upper-triangular]] with the same [[diagonal]] as $H$: $H ^{-1}=\begin{bmatrix} 1 & -* & *^{2}-* & \\ 0 & 1 & -* \\ 0 & 0 & 1 \end{bmatrix}.$Thus the subset $H$ is closed under inverses. It is closed under [[matrix product|matrix multiplication]] too, for letting $H_{1}$ and $H_{2}$ in $H$ we can compute $H_{1}H_{2}=\begin{bmatrix} 1 & \star & \star \\ & 1 & \star \\ & & 1 \end{bmatrix}\begin{bmatrix} 1 & @ & @ \\ & 1 & @ \\ & & 1 \end{bmatrix} = \begin{bmatrix} 1 & @+\star & @+\star @ + \star \\ 0 & 1 & @ + 1 \\ 0 & 0 & 1 \end{bmatrix}.$ Of course, $I_{3} \in H$ as well; thus $H$ is a [[subgroup]] of $GL_{3}(\mathbb{R})$. \ To show that $K$ is a [[normal subgroup]] of $H$, we need to show that for any $h \in H$ we have that $hKh ^{-1}$ takes on the form $\begin{bmatrix} 1 & 0 & * \\ & 1 & 0 \\ & & 1 \end{bmatrix}.$ $H$ is normal by the following computation. ![[CleanShot 2023-09-17 at 20.10.57.jpg]] ![[CleanShot 2023-09-17 at 20.12.18.jpg]] ^7aaa6f > [!warning] > If $N_{1} \trianglelefteq G_{1}$ and $N_{2} \trianglelefteq G_{2}$ with $G_{1} / N_{1} \cong G_{2} / N_{2}$, it is not generally true that $G_{1} \cong C_{2}$. A counterexample can be obtained by considering that $D_{4} / \langle x \rangle \cong Q_{8} / \langle \v k \rangle\cong C_{2}$, but $D_{4} \not \cong Q_{8}$. > \ > Similarly, if $N_{1} \trianglelefteq G_{1}$ and $N_{2} \trianglelefteq G_{2}$ with $G_{1} \cong G_{2}$ and $G_{1} / N_{1} \cong G_{2} / N_{2}$, then it is not generally true that $N_{1} \cong N_{2}$. For example $D_{4}$ has a [[subgroup]] $\langle x \rangle \cong C_{4}$ as well as a [[subgroup]] $\{ e,y,x^{2}, yx^{2} \} \cong C_{2} \times C_{2}$. Both these groups have order $4$, so quotienting $D_{4}$ by them yields $C_{2}$. But they are not isomorphic. > \ > > [!basicexample] [[Subgroup]]s of $D_{4} / Z(D_{4})$ > ![[CleanShot 2023-09-18 at 19.07.54.jpg]] > ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```