---- > [!definition] Definition. ([[normal subgroup]]) > A [[subgroup]] $N$ of a [[group]] $G$ is **normal** if for all $g\in G$, the left $N$-[[coset]] $gN$and the right $N$-[[coset|coset]] $Ng$ are the *same* subset of $G$, i.e., $gN=Ng \text{ for all } g \in G.$ > > > We write $N \trianglelefteq G$. > [!equivalence] > ![[conjugate characterization of normal subgroups#^53879a]] ^dc9015 ^5bd15e > [!warning] > Keep in mind that gN=Ng'$ does not mean $g$ commutes with every element in $N$. It only means that, fixing $n \in N$ and considering $gn$, that there exists some $n' \in N$ such that $gn=n'g$. (to have $gN \subset Ng$, and then the other direction for the reverse inclusion) > [!warning] > If $H \trianglelefteq N$ and $N \trianglelefteq G$, we cannot say $H \trianglelefteq G$! For example, consider the [[dihedral group]] $D_{4}$. The [[subgroup]] $\{ e,y,yx^{2},x^{2} \}$ is [[normal subgroup|normal]] in $D_{4}$ ([[subgroups of index 2 are normal|index 2]]), and its subgroup $\{ e,y \}$ is normal in it. But $\{ e,y \}$ is not normal in $D_{4}$ (e.g., $x\{ e,y \}=\{x,yx^{2}\}\neq \{ e,y \}x=\{ x,yx \}.$) >[!basicexample] > - [[kernel iff normal subgroup]] > - [[subgroups of abelian groups are normal]] > - [[subgroups of index 2 are normal]] > - ![[CleanShot 2023-09-10 at [email protected]]] > - ![[CleanShot 2023-09-10 at 12.01.01.jpg]] > - ![[CleanShot 2023-09-10 at [email protected]]] > - The [[special linear group]] is a [[normal subgroup]] of the [[general linear group]], since [[change of basis formula|changing basis]] does not change the [[determinant of a matrix]] ([[eigenvalue]]s don't change so neither do their product). > - Recall the [[dihedral group#^d81188|subgroups of]] $D_{4}$. Of these, the [[normal subgroup]]s of $D_{4}$ are: **Order $4$ subgroups[^1]:** $\{ e, x^{2},yx,yx^{3} \}, \{ e, x^{2}, x^{3},x^{4} \}, \{ e , x^{2}, y, yx^{2} \}$; $\{ e, x^{2} \}$: Invariance under [[conjugate|conjugation]] is easy to check. $\{ e \}$: trivial. \ We can find counterexamples for the other [[subgroup]]s (just need one non-invariant [[conjugate]]). > [!basicproperties] > - A [[subgroup]] $\{ e,a \}$ of [[order of a group|order]] $2$ is [[normal subgroup|normal]] only when it is contained in the [[center of a group|group's center]], since we need $ag=ga$ for all $g \in G$. [^1]: This is due to [[Lagrange's Theorem]] and associated **lemmas**: let $H \leq D_{4}$ have order $4$. All [[coset]]s of $H$ have equal [[cardinality]]; $|eH|$ is a [[coset]] so this [[cardinality]] is in fact $|H|$. Since the [[coset]]s of $H$ [[partition]] $D_{4}$ and $H$ is already a left and right [[coset]], there can only exist one other left [[coset]] $H$ and one other right coset of $H$, and they must be equal. ---- #### ---- #### 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 > ```