Normal subgroup

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 ${\displaystyle N}${\displaystyle N} of the group ${\displaystyle G}${\displaystyle G} is normal in ${\displaystyle G}${\displaystyle G} if and only if ${\displaystyle gng^{-1}\in N}${\displaystyle gng^{-1}\in N} for all ${\displaystyle g\in G}${\displaystyle g\in G} and ${\displaystyle n\in N}${\displaystyle n\in N}. The usual notation for this relation is ${\displaystyle N\triangleleft G}${\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 ${\displaystyle G}${\displaystyle G} are precisely the kernels of group homomorphisms with domain ${\displaystyle G}${\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]

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

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

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

For any group ${\displaystyle G}${\displaystyle G}, the trivial subgroup ${\displaystyle \{e\}}${\displaystyle \{e\}} consisting of only the identity element of ${\displaystyle G}${\displaystyle G} is always a normal subgroup of ${\displaystyle G}${\displaystyle G}. Likewise, ${\displaystyle G}${\displaystyle G} itself is always a normal subgroup of ${\displaystyle G}${\displaystyle G} (if these are the only normal subgroups, then ${\displaystyle G}${\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 ${\displaystyle [G,G]}${\displaystyle [G,G]}.^{[7] [8]} More generally, since conjugation is an isomorphism, any characteristic subgroup is a normal subgroup.^[9]

If ${\displaystyle G}${\displaystyle G} is an abelian group then every subgroup ${\displaystyle N}${\displaystyle N} of ${\displaystyle G}${\displaystyle G} is normal, because ${\displaystyle gN=\{gn\}_{n\in N}=\{ng\}_{n\in N}=Ng}${\displaystyle gN=\{gn\}_{n\in N}=\{ng\}_{n\in N}=Ng}. More generally, for any group ${\displaystyle G}${\displaystyle G}, every subgroup of the center ${\displaystyle Z(G)}${\displaystyle Z(G)} of ${\displaystyle G}${\displaystyle G} is normal in ${\displaystyle G}${\displaystyle G} (in the special case that ${\displaystyle G}${\displaystyle G} is abelian, the center is all of ${\displaystyle G}${\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 a Hamiltonian group.^[10]

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

As an example of a normal subgroup within a matrix group, consider the general linear group ${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )} of all invertible ${\displaystyle n\times n}${\displaystyle n\times n} matrices with real entries under the operation of matrix multiplication and its subgroup ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )} of all ${\displaystyle n\times n}${\displaystyle n\times n} matrices of determinant 1 (the special linear group). To see why the subgroup ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )} is normal in ${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}, consider any matrix ${\displaystyle X}${\displaystyle X} in ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )} and any invertible matrix ${\displaystyle A}${\displaystyle A}. Then using the two important identities ${\displaystyle \det(AB)=\det(A)\det(B)}${\displaystyle \det(AB)=\det(A)\det(B)} and ${\displaystyle \det(A^{-1})=\det(A)^{-1}}${\displaystyle \det(A^{-1})=\det(A)^{-1}}, one has that ${\displaystyle \det(AXA^{-1})=\det(A)\det(X)\det(A)^{-1}=\det(X)=1}${\displaystyle \det(AXA^{-1})=\det(A)\det(X)\det(A)^{-1}=\det(X)=1}, and so ${\displaystyle AXA^{-1}\in \mathrm {SL} _{n}(\mathbf {R} )}${\displaystyle AXA^{-1}\in \mathrm {SL} _{n}(\mathbf {R} )} as well. This means ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )} is closed under conjugation in ${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}, so it is a normal subgroup.^[a]

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.

• If ${\displaystyle H}${\displaystyle H} is a normal subgroup of ${\displaystyle G}${\displaystyle G}, and ${\displaystyle K}${\displaystyle K} is a subgroup of ${\displaystyle G}${\displaystyle G} containing ${\displaystyle H}${\displaystyle H}, then ${\displaystyle H}${\displaystyle H} is a normal subgroup of ${\displaystyle K}${\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 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 ${\displaystyle G}${\displaystyle G} and ${\displaystyle H}${\displaystyle H} are normal subgroups of their direct product ${\displaystyle G\times H}${\displaystyle G\times H}.
• If the group ${\displaystyle G}${\displaystyle G} is a semidirect product ${\displaystyle G=N\rtimes H}${\displaystyle G=N\rtimes H}, then ${\displaystyle N}${\displaystyle N} is normal in ${\displaystyle G}${\displaystyle G}, though ${\displaystyle H}${\displaystyle H} need not be normal in ${\displaystyle G}${\displaystyle G}.
• If ${\displaystyle M}${\displaystyle M} and ${\displaystyle N}${\displaystyle N} are normal subgroups of an additive group ${\displaystyle G}${\displaystyle G} such that ${\displaystyle G=M+N}${\displaystyle G=M+N} and ${\displaystyle M\cap N=\{0\}}${\displaystyle M\cap N=\{0\}}, then ${\displaystyle G=M\oplus N}${\displaystyle G=M\oplus N}.^[18]
• Normality is preserved under surjective homomorphisms;^[19] that is, if ${\displaystyle G\to H}${\displaystyle G\to H} is a surjective group homomorphism and ${\displaystyle N}${\displaystyle N} is normal in ${\displaystyle G}${\displaystyle G}, then the image ${\displaystyle f(N)}${\displaystyle f(N)} is normal in ${\displaystyle H}${\displaystyle H}.
• Normality is preserved by taking inverse images;^[19] that is, if ${\displaystyle G\to H}${\displaystyle G\to H} is a group homomorphism and ${\displaystyle N}${\displaystyle N} is normal in ${\displaystyle H}${\displaystyle H}, then the inverse image ${\displaystyle f^{-1}(N)}${\displaystyle f^{-1}(N)} is normal in ${\displaystyle G}${\displaystyle G}.
• Normality is preserved on taking direct products;^[20] that is, if ${\displaystyle N_{1}\triangleleft G_{1}}${\displaystyle N_{1}\triangleleft G_{1}} and ${\displaystyle N_{2}\triangleleft G_{2}}${\displaystyle N_{2}\triangleleft G_{2}}, then ${\displaystyle N_{1}\times N_{2}\;\triangleleft \;G_{1}\times G_{2}}${\displaystyle N_{1}\times N_{2}\;\triangleleft \;G_{1}\times G_{2}}.
• Every subgroup of index 2 is normal. More generally, a subgroup, ${\displaystyle H}${\displaystyle H}, of finite index, ${\displaystyle n}${\displaystyle n}, in ${\displaystyle G}${\displaystyle G} contains a subgroup, ${\displaystyle K,}${\displaystyle K,} normal in ${\displaystyle G}${\displaystyle G} and of index dividing ${\displaystyle n!}${\displaystyle n!} called the normal core. In particular, if ${\displaystyle p}${\displaystyle p} is the smallest prime dividing the order of ${\displaystyle G}${\displaystyle G}, then every subgroup of index ${\displaystyle p}${\displaystyle p} is normal.^[21]
• The fact that normal subgroups of ${\displaystyle G}${\displaystyle G} are precisely the kernels of group homomorphisms defined on ${\displaystyle G}${\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 is simple if and only if it is isomorphic to all of its non-identity homomorphic images,^[22] 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.

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

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

The lattice is complete and modular.^[20]

If ${\displaystyle N}${\displaystyle N} is a normal subgroup, we can define a multiplication on cosets as follows: $${\displaystyle \left(a_{1}N\right)\left(a_{2}N\right):=\left(a_{1}a_{2}\right)N}$${\displaystyle \left(a_{1}N\right)\left(a_{2}N\right):=\left(a_{1}a_{2}\right)N} This relation defines a mapping ${\displaystyle G/N\times G/N\to G/N}${\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 ${\displaystyle a_{1},a_{2}}${\displaystyle a_{1},a_{2}} does not affect the result. To this end, consider some other representative elements ${\displaystyle a_{1}'\in a_{1}N,a_{2}'\in a_{2}N}${\displaystyle a_{1}'\in a_{1}N,a_{2}'\in a_{2}N}. Then there are ${\displaystyle n_{1},n_{2}\in N}${\displaystyle n_{1},n_{2}\in N} such that ${\displaystyle a_{1}'=a_{1}n_{1},a_{2}'=a_{2}n_{2}}${\displaystyle a_{1}'=a_{1}n_{1},a_{2}'=a_{2}n_{2}}. It follows that $${\displaystyle a_{1}'a_{2}'N=a_{1}n_{1}a_{2}n_{2}N=a_{1}a_{2}n_{1}'n_{2}N=a_{1}a_{2}N}$${\displaystyle 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 ${\displaystyle N}${\displaystyle N} is a normal subgroup, and therefore there is ${\displaystyle n_{1}'\in N}${\displaystyle n_{1}'\in N} such that ${\displaystyle n_{1}a_{2}=a_{2}n_{1}'}${\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 ${\displaystyle G/N.}${\displaystyle G/N.} There is a natural homomorphism, ${\displaystyle f:G\to G/N}${\displaystyle f:G\to G/N}, given by ${\displaystyle f(a)=aN}${\displaystyle f(a)=aN}. This homomorphism maps ${\displaystyle N}${\displaystyle N} into the identity element of ${\displaystyle G/N}${\displaystyle G/N}, which is the coset ${\displaystyle eN=N}${\displaystyle eN=N},^[23] that is, ${\displaystyle \ker(f)=N}${\displaystyle \ker(f)=N}.

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

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

• 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?

• Subgroup properties

• Articles with short description
• Short description matches Wikidata
• All articles with unsourced statements
• Articles with unsourced statements from March 2019
• Articles with unsourced statements from October 2020