Normal subgroup

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In abstract algebra, a normal subgroup (also known as an invariant subgroup or self-conjugate subgroup)^[1] is a subgroup that is invariant under conjugation by members of the group of which it is a part. In other words, a subgroup ${\displaystyle N}$ of the group ${\displaystyle G}$ is normal in ${\displaystyle G}$ if and only if ${\displaystyle gng^{-1}\in N}$ for all ${\displaystyle g\in G}$ and ${\displaystyle n\in N.}$ The usual notation for this relation is ${\displaystyle N\triangleleft G.}$

Normal subgroups are important because they (and only they) can be used to construct quotient groups of the given group. Furthermore, the normal subgroups of ${\displaystyle G}$ are precisely the kernels of group homomorphisms with domain ${\displaystyle G,}$ 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[]

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

Equivalent conditions[]

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

• The image of conjugation of ${\displaystyle N}$ by any element of ${\displaystyle G}$ is a subset of ${\displaystyle N.}$^[4]
• The image of conjugation of ${\displaystyle N}$ by any element of ${\displaystyle G}$ is equal to ${\displaystyle N.}$^[4]
• For all ${\displaystyle g\in G,}$ the left and right cosets ${\displaystyle gN}$ and ${\displaystyle Ng}$ are equal.^[4]
• The sets of left and right cosets of ${\displaystyle N}$ in ${\displaystyle G}$ coincide.^[4]
• The product of an element of the left coset of ${\displaystyle N}$ with respect to ${\displaystyle g}$ and an element of the left coset of ${\displaystyle N}$ with respect to ${\displaystyle h}$ is an element of the left coset of ${\displaystyle N}$ with respect to ${\displaystyle gh}$: for all ${\displaystyle x,y,g,h\in G,}$ if ${\displaystyle x\in gN}$and ${\displaystyle y\in hN}$ then ${\displaystyle xy\in (gh)N.}$
• ${\displaystyle N}$ is a union of conjugacy classes of ${\displaystyle G.}$^[2]
• ${\displaystyle N}$ is preserved by the inner automorphisms of ${\displaystyle G.}$^[5]
• There is some group homomorphism ${\displaystyle G\to H}$ whose kernel is ${\displaystyle N.}$^[2]
• For all ${\displaystyle n\in N}$ and ${\displaystyle g\in G,}$ the commutator ${\displaystyle [n,g]=n^{-1}g^{-1}ng}$ is in ${\displaystyle N.}$^{[citation needed]}
• Any two elements commute regarding the normal subgroup membership relation: for all ${\displaystyle g,h\in G,}$ ${\displaystyle gh\in N}$ if and only if ${\displaystyle hg\in N.}$^{[citation needed]}

Examples[]

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

If ${\displaystyle G}$ is an abelian group then every subgroup ${\displaystyle N}$ of ${\displaystyle G}$ is normal, because ${\displaystyle gN=\{gn\}_{n\in N}=\{ng\}_{n\in N}=Ng.}$ A group that is not abelian but for which every subgroup is normal is called a Hamiltonian group.^[10]

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

In the Rubik'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]

The translation group is a normal subgroup of the Euclidean 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 all rotations about the origin is not 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[]

• If ${\displaystyle H}$ is a normal subgroup of ${\displaystyle G,}$ and ${\displaystyle K}$ is a subgroup of ${\displaystyle G}$ containing ${\displaystyle H,}$ then ${\displaystyle H}$ is a normal subgroup of ${\displaystyle K.}$^[14]
• A normal subgroup of a normal subgroup of a group need not be normal in the group. That is, normality is not a transitive relation. The smallest group exhibiting this phenomenon is the dihedral group of order 8.^[15] However, a characteristic subgroup of a normal subgroup is normal.^[16] A group in which normality is transitive is called a T-group.^[17]
• The two groups ${\displaystyle G}$ and ${\displaystyle H}$ are normal subgroups of their direct product ${\displaystyle G\times H.}$
• If the group ${\displaystyle G}$ is a semidirect product ${\displaystyle G=N\rtimes H,}$ then ${\displaystyle N}$ is normal in ${\displaystyle G,}$ though ${\displaystyle H}$ need not be normal in ${\displaystyle G.}$
• Normality is preserved under surjective homomorphisms;^[18] that is, if ${\displaystyle G\to H}$ is a surjective group homomorphism and ${\displaystyle N}$ is normal in ${\displaystyle G,}$ then the image ${\displaystyle f(N)}$ is normal in ${\displaystyle H.}$
• Normality is preserved by taking inverse images;^[18] that is, if ${\displaystyle G\to H}$ is a group homomorphism and ${\displaystyle N}$ is normal in ${\displaystyle H,}$ then the inverse image ${\displaystyle f^{-1}(N)}$ is normal in ${\displaystyle G.}$
• Normality is preserved on taking direct products;^[19] that is, if ${\displaystyle N_{1}\triangleleft G_{1}}$ and ${\displaystyle N_{2}\triangleleft G_{2},}$ then ${\displaystyle N_{1}\times N_{2}\;\triangleleft \;G_{1}\times G_{2}.}$
• Every subgroup of index 2 is normal. More generally, a subgroup, ${\displaystyle H,}$ of finite index, ${\displaystyle n,}$ in ${\displaystyle G}$ contains a subgroup, ${\displaystyle K,}$ normal in ${\displaystyle G}$ and of index dividing ${\displaystyle n!}$ called the normal core. In particular, if ${\displaystyle p}$ is the smallest prime dividing the order of ${\displaystyle G,}$ then every subgroup of index ${\displaystyle p}$ is normal.^[20]
• The fact that normal subgroups of ${\displaystyle G}$ are precisely the kernels of group homomorphisms defined on ${\displaystyle G}$ 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 is simple if and only if it is isomorphic to all of its non-identity homomorphic images,^[21] a finite group is perfect if and only if it has no normal subgroups of prime index, and a group is imperfect if and only if the derived subgroup is not supplemented by any proper normal subgroup.

Lattice of normal subgroups[]

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

The normal subgroups of ${\displaystyle G}$ form a lattice under subset inclusion with least element, ${\displaystyle \{e\},}$ and greatest element, ${\displaystyle G.}$ The meet of two normal subgroups, ${\displaystyle N}$ and ${\displaystyle M,}$ in this lattice is their intersection and the join is their product.

The lattice is complete and modular.^[19]

Normal subgroups, quotient groups and homomorphisms[]

If ${\displaystyle N}$ is a normal subgroup, we can define a multiplication on cosets as follows:

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

With this operation, the set of cosets is itself a group, called the quotient group and denoted with ${\displaystyle G/N.}$ There is a natural homomorphism, ${\displaystyle f:G\to G/N,}$ given by ${\displaystyle f(a)=aN.}$ This homomorphism maps ${\displaystyle N}$ into the identity element of ${\displaystyle G/N,}$ which is the coset ${\displaystyle eN=N,}$^[22] that is, ${\displaystyle \ker(f)=N.}$

In general, a group homomorphism, ${\displaystyle f:G\to H}$ sends subgroups of ${\displaystyle G}$ to subgroups of ${\displaystyle H.}$ Also, the preimage of any subgroup of ${\displaystyle H}$ is a subgroup of ${\displaystyle G.}$ We call the preimage of the trivial group ${\displaystyle \{e\}}$ in ${\displaystyle H}$ the kernel of the homomorphism and denote it by ${\displaystyle \ker f.}$ As it turns out, the kernel is always normal and the image of ${\displaystyle G,f(G),}$ is always isomorphic to ${\displaystyle G/\ker f}$ (the first isomorphism theorem).^[23] In fact, this correspondence is a bijection between the set of all quotient groups of ${\displaystyle G,G/N,}$ and the set of all homomorphic images of ${\displaystyle G}$ (up to isomorphism).^[24] It is also easy to see that the kernel of the quotient map, ${\displaystyle f:G\to G/N,}$ is ${\displaystyle N}$ itself, so the normal subgroups are precisely the kernels of homomorphisms with domain ${\displaystyle G.}$^[25]

See also[]

Notes[]

References[]

Further reading[]

• I. N. Herstein, Topics in algebra. Second edition. Xerox College Publishing, Lexington, Mass.-Toronto, Ont., 1975. xi+388 pp.

External links[]

• Weisstein, Eric W. "normal subgroup". MathWorld.
• Normal subgroup in Springer's Encyclopedia of Mathematics
• Robert Ash: Group Fundamentals in Abstract Algebra. The Basic Graduate Year
• Timothy Gowers, Normal subgroups and quotient groups
• John Baez, What's a Normal Subgroup?