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Normal subgroup

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Subgroup invariant under conjugation

"Invariant subgroup" redirects here; not to be confused withFully invariant subgroup.

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Inabstract algebra, anormal subgroup (also known as aninvariant subgroup orself-conjugate subgroup)^[1] is asubgroup that isinvariant underconjugation by members of thegroup of which it is a part. In other words, a subgroup $N {\displaystyle N}$ of the group $G {\displaystyle G}$ is normal in $G {\displaystyle G}$ if and only if $gng^{-1}\in N$ for all $g\in G$ and $n\in N.$ The usual notation for this relation is $N\triangleleft G.$

Normal subgroups are important because they (and only they) can be used to constructquotient groups of the given group. Furthermore, the normal subgroups of $G {\displaystyle G}$ are precisely thekernels ofgroup homomorphisms withdomain $G, {\displaystyle G,}$ which means that they can be used to internally classify those homomorphisms.

Évariste Galois was the first to realize the importance of the existence of normal subgroups.^[2]

Definitions

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Asubgroup $N {\displaystyle N}$ of a group $G {\displaystyle G}$ is called anormal subgroup of $G {\displaystyle G}$ if it is invariant underconjugation; that is, the conjugation of an element of $N {\displaystyle N}$ by an element of $G {\displaystyle G}$ is always in $N . {\displaystyle N.}$ ^[3] The usual notation for this relation is $N\triangleleft G.$

Equivalent conditions

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For any subgroup $N {\displaystyle N}$ of $G, {\displaystyle G,}$ the following conditions areequivalent to $N {\displaystyle N}$ being a normal subgroup of $G . {\displaystyle G.}$ Therefore, any one of them may be taken as the definition.

The image of conjugation of $N {\displaystyle N}$ by any element of $G {\displaystyle G}$ is a subset of $N, {\displaystyle N,}$ ^[4] i.e., $gNg^{-1}\subseteq N$ for all $g\in G$ .
The image of conjugation of $N {\displaystyle N}$ by any element of $G {\displaystyle G}$ is equal to $N, {\displaystyle N,}$ ^[4] i.e., $gNg^{-1}=N$ for all $g\in G$ .
For all $g\in G,$ the left and rightcosets $g N {\displaystyle gN}$ and $N g {\displaystyle Ng}$ are equal.^[4]
The sets of left and rightcosets of $N {\displaystyle N}$ in $G {\displaystyle G}$ coincide.^[4]
Multiplication in $G {\displaystyle G}$ preserves the equivalence relation "is in the same left coset as". That is, for every $g,g',h,h'\in G$ satisfying $gN=g'N$ and $hN=h'N$ , we have $(gh)N=(g'h')N.$
There exists a group on the set of left cosets of $N {\displaystyle N}$ where multiplication of any two left cosets $g N {\displaystyle gN}$ and $h N {\displaystyle hN}$ yields the left coset $(gh)N$ . (This group is called thequotient group of $G {\displaystyle G}$ modulo $N {\displaystyle N}$ , denoted $G/N$ .)
$N {\displaystyle N}$ is aunion ofconjugacy classes of $G . {\displaystyle G.}$ ^[2]
$N {\displaystyle N}$ is preserved by theinner automorphisms of $G . {\displaystyle G.}$ ^[5]
There is somegroup homomorphism $G\to H$ whosekernel is $N . {\displaystyle N.}$ ^[2]
There exists a group homomorphism $\phi :G\to H$ whosefibers form a group where the identity element is $N {\displaystyle N}$ and multiplication of any two fibers $\phi ^{-1}(h_{1})$ and $\phi ^{-1}(h_{2})$ yields the fiber $\phi ^{-1}(h_{1}h_{2})$ . (This group is the same group $G/N$ mentioned above.)
There is somecongruence relation on $G {\displaystyle G}$ for which theequivalence class of theidentity element is $N {\displaystyle N}$ .
For all $n\in N$ and $g\in G,$ thecommutator $[n,g]=n^{-1}g^{-1}ng$ is in $N . {\displaystyle N.}$ ^{[citation needed]}
Any two elements commute modulo the normal subgroup membership relation. That is, for all $g,h\in G,$ $gh\in N$ if and only if $hg\in N.$ ^{[citation needed]}

Examples

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For any group $G, {\displaystyle G,}$ the trivial subgroup $\{e\}$ consisting of just the identity element of $G {\displaystyle G}$ is always a normal subgroup of $G . {\displaystyle G.}$ Likewise, $G {\displaystyle G}$ itself is always a normal subgroup of $G . {\displaystyle G.}$ (If these are the only normal subgroups, then $G {\displaystyle G}$ is said to besimple.)^[6] Other named normal subgroups of an arbitrary group include thecenter of the group (the set of elements that commute with all other elements) and thecommutator subgroup $[G,G].$ ^[7]^[8] More generally, since conjugation is an isomorphism, anycharacteristic subgroup is a normal subgroup.^[9]

If $G {\displaystyle G}$ is anabelian group then every subgroup $N {\displaystyle N}$ of $G {\displaystyle G}$ is normal, because $gN=\{gn\}_{n\in N}=\{ng\}_{n\in N}=Ng.$ More generally, for any group $G {\displaystyle G}$ , every subgroup of thecenter $Z(G)$ of $G {\displaystyle G}$ is normal in $G {\displaystyle G}$ . (In the special case that $G {\displaystyle G}$ is abelian, the center is all of $G {\displaystyle G}$ , hence the fact that all subgroups of an abelian group are normal.) A group that is not abelian but for which every subgroup is normal is called aHamiltonian group.^[10]

A concrete example of a normal subgroup is the subgroup $N=\{(1),(123),(132)\}$ of thesymmetric group $S_{3},$ consisting of the identity and both three-cycles. In particular, one can check that every coset of $N {\displaystyle N}$ is either equal to $N {\displaystyle N}$ itself or is equal to $(12)N=\{(12),(23),(13)\}.$ On the other hand, the subgroup $H=\{(1),(12)\}$ is not normal in $S_{3}$ since $(123)H=\{(123),(13)\}\neq \{(123),(23)\}=H(123).$ ^[11] This illustrates the general fact that any subgroup $H\leq G$ ofindex two is normal.

In theRubik's Cube group, the subgroups consisting of operations which only affect the orientations of either the corner pieces or the edge pieces are normal.^[12]

Thetranslation group is a normal subgroup of theEuclidean group in any dimension.^[13] This means: applying a rigid transformation, followed by a translation and then the inverse rigid transformation, has the same effect as a single translation. By contrast, the subgroup of allrotations about the origin isnot a normal subgroup of the Euclidean group, as long as the dimension is at least 2: first translating, then rotating about the origin, and then translating back will typically not fix the origin and will therefore not have the same effect as a single rotation about the origin.

Properties

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If $H {\displaystyle H}$ is a normal subgroup of $G, {\displaystyle G,}$ and $K {\displaystyle K}$ is a subgroup of $G {\displaystyle G}$ containing $H, {\displaystyle H,}$ then $H {\displaystyle H}$ is a normal subgroup of $K . {\displaystyle K.}$ ^[14]
A normal subgroup of a normal subgroup of a group need not be normal in the group. That is, normality is not atransitive relation. The smallest group exhibiting this phenomenon is thedihedral group of order 8.^[15] However, acharacteristic subgroup of a normal subgroup is normal.^[16] A group in which normality is transitive is called aT-group.^[17]
The two groups $G {\displaystyle G}$ and $H {\displaystyle H}$ are normal subgroups of theirdirect product $G\times H.$
If the group $G {\displaystyle G}$ is asemidirect product $G=N\rtimes H,$ then $N {\displaystyle N}$ is normal in $G, {\displaystyle G,}$ though $H {\displaystyle H}$ need not be normal in $G . {\displaystyle G.}$
If $M {\displaystyle M}$ and $N {\displaystyle N}$ are normal subgroups of an additive group $G {\displaystyle G}$ such that $G=M+N$ and $M\cap N=\{0\}$ , then $G=M\oplus N.$ ^[18]
Normality is preserved under surjective homomorphisms;^[19] that is, if $G\to H$ is a surjective group homomorphism and $N {\displaystyle N}$ is normal in $G, {\displaystyle G,}$ then the image $f(N)$ is normal in $H . {\displaystyle H.}$
Normality is preserved by takinginverse images;^[19] that is, if $G\to H$ is a group homomorphism and $N {\displaystyle N}$ is normal in $H, {\displaystyle H,}$ then the inverse image $f^{-1}(N)$ is normal in $G . {\displaystyle G.}$
Normality is preserved on takingdirect products;^[20] that is, if $N_{1}\triangleleft G_{1}$ and $N_{2}\triangleleft G_{2},$ then $N_{1}\times N_{2}\;\triangleleft \;G_{1}\times G_{2}.$
Every subgroup ofindex 2 is normal. More generally, a subgroup, $H, {\displaystyle H,}$ of finite index, $n, {\displaystyle n,}$ in $G {\displaystyle G}$ contains a subgroup, $K, {\displaystyle K,}$ normal in $G {\displaystyle G}$ and of index dividing $n! {\displaystyle n!}$ called thenormal core. In particular, if $p {\displaystyle p}$ is the smallest prime dividing the order of $G, {\displaystyle G,}$ then every subgroup of index $p {\displaystyle p}$ is normal.^[21]
The fact that normal subgroups of $G {\displaystyle G}$ are precisely the kernels of group homomorphisms defined on $G {\displaystyle G}$ accounts for some of the importance of normal subgroups; they are a way to internally classify all homomorphisms defined on a group. For example, a non-identity finite group issimple if and only if it is isomorphic to all of its non-identity homomorphic images,^[22] a finite group isperfect if and only if it has no normal subgroups of primeindex, and a group isimperfect if and only if thederived subgroup is not supplemented by any proper normal subgroup.

Lattice of normal subgroups

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Given two normal subgroups, $N {\displaystyle N}$ and $M, {\displaystyle M,}$ of $G, {\displaystyle G,}$ their intersection $N\cap M$ and theirproduct $NM=\{nm:n\in N\;{\text{ and }}\;m\in M\}$ are also normal subgroups of $G . {\displaystyle G.}$

The normal subgroups of $G {\displaystyle G}$ form alattice undersubset inclusion withleast element, $\{e\},$ andgreatest element, $G . {\displaystyle G.}$ Themeet of two normal subgroups, $N {\displaystyle N}$ and $M, {\displaystyle M,}$ in this lattice is their intersection and thejoin is their product.

The lattice iscomplete andmodular.^[20]

Normal subgroups, quotient groups and homomorphisms

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If $N {\displaystyle N}$ is a normal subgroup, we can define a multiplication on cosets as follows: $\left(a_{1}N\right)\left(a_{2}N\right):=\left(a_{1}a_{2}\right)N.$ This relation defines a mapping $G/N\times G/N\to G/N.$ To show that this mapping is well-defined, one needs to prove that the choice of representative elements $a_{1},a_{2}$ does not affect the result. To this end, consider some other representative elements $a_{1}'\in a_{1}N,a_{2}'\in a_{2}N.$ Then there are $n_{1},n_{2}\in N$ such that $a_{1}'=a_{1}n_{1},a_{2}'=a_{2}n_{2}.$ It follows that $a_{1}'a_{2}'N=a_{1}n_{1}a_{2}n_{2}N=a_{1}a_{2}n_{1}'n_{2}N=a_{1}a_{2}N,$ where we also used the fact that $N {\displaystyle N}$ is anormal subgroup, and therefore there is $n_{1}'\in N$ such that $n_{1}a_{2}=a_{2}n_{1}'.$ This proves that this product is a well-defined mapping between cosets.

With this operation, the set of cosets is itself a group, called thequotient group and denoted with $G/N.$ There is a naturalhomomorphism, $f:G\to G/N,$ given by $f(a)=aN.$ This homomorphism maps $N {\displaystyle N}$ into the identity element of $G/N,$ which is the coset $eN=N,$ ^[23] that is, $\ker(f)=N.$

In general, a group homomorphism, $f:G\to H$ sends subgroups of $G {\displaystyle G}$ to subgroups of $H . {\displaystyle H.}$ Also, the preimage of any subgroup of $H {\displaystyle H}$ is a subgroup of $G . {\displaystyle G.}$ We call the preimage of the trivial group $\{e\}$ in $H {\displaystyle H}$ thekernel of the homomorphism and denote it by $\ker f.$ As it turns out, the kernel is always normal and the image of $G,f(G),$ is alwaysisomorphic to $G/\ker f$ (thefirst isomorphism theorem).^[24] In fact, this correspondence is a bijection between the set of all quotient groups of $G,G/N,$ and the set of all homomorphic images of $G {\displaystyle G}$ (up to isomorphism).^[25] It is also easy to see that the kernel of the quotient map, $f:G\to G/N,$ is $N {\displaystyle N}$ itself, so the normal subgroups are precisely the kernels of homomorphisms withdomain $G . {\displaystyle G.}$ ^[26]

Notes

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^In other language: $\det$ is a homomorphism from $\mathrm {GL} _{n}(\mathbf {R} )$ to the multiplicative subgroup $\mathbf {R} ^{\times }$ , and $\mathrm {SL} _{n}(\mathbf {R} )$ is the kernel. Both arguments also work over thecomplex numbers, or indeed over an arbitraryfield.

References

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Bibliography

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