Normal subgroup

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 ${\displaystyle N\triangleleft G.}$

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

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 ${\displaystyle gN=\{gn\}_{n\in N}=\{ng\}_{n\in N}=Ng.}$ More generally, for any group ${\displaystyle G}$, every subgroup of the center ${\displaystyle Z(G)}$ of ${\displaystyle G}$ is normal in ${\displaystyle G}$. (In the special case that ${\displaystyle G}$ is abelian, the center is all of ${\displaystyle G}$, 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 ${\displaystyle N=\{(1),(123),(132)\}}$ 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] This illustrates the general fact that any subgroup ${\displaystyle H\leq G}$ of index two is normal.

As an example of a normal subgroup within a matrix group, consider the general linear group ${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}$ of all invertible ${\displaystyle n\times n}$ matrices with real entries under the operation of matrix multiplication and its subgroup ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}$ of all ${\displaystyle n\times n}$ matrices of determinant 1 (the special linear group). To see why the subgroup ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}$ is normal in ${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}$, consider any matrix ${\displaystyle X}$ in ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}$ and any invertible matrix ${\displaystyle A}$. Then using the two important identities ${\displaystyle \det(AB)=\det(A)\det(B)}$ and ${\displaystyle \det(A^{-1})=\det(A)^{-1}}$, one has that ${\displaystyle \det(AXA^{-1})=\det(A)\det(X)\det(A)^{-1}=\det(X)=1}$, and so ${\displaystyle AXA^{-1}\in \mathrm {SL} _{n}(\mathbf {R} )}$ as well. This means ${\displaystyle \mathrm {SL} _{n}(\mathbf {R} )}$ is closed under conjugation in ${\displaystyle \mathrm {GL} _{n}(\mathbf {R} )}$, 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 ${\displaystyle H}$ 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 ${\displaystyle G.}$
• If ${\displaystyle M}$ and ${\displaystyle N}$ are normal subgroups of an additive group ${\displaystyle G}$ such that ${\displaystyle G=M+N}$ and ${\displaystyle M\cap N=\{0\}}$, then ${\displaystyle G=M\oplus N.}$^[18]
• Normality is preserved under surjective homomorphisms;^[19] 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;^[19] 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;^[20] 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.^[21]
• 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,^[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, ${\displaystyle N}$ 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.^[20]

If ${\displaystyle N}$ is a normal subgroup, we can define a multiplication on cosets as follows: $${\displaystyle \left(a_{1}N\right)\left(a_{2}N\right):=\left(a_{1}a_{2}\right)N.}$$ 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 $${\displaystyle a_{1}'a_{2}'N=a_{1}n_{1}a_{2}n_{2}N=a_{1}a_{2}n_{1}'n_{2}N=a_{1}a_{2}N,}$$where we also used the fact that ${\displaystyle N}$ 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 ${\displaystyle eN=N,}$^[23] 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).^[24] 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).^[25] 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 ${\displaystyle G.}$^[26]