Normal subgroup

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.) 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 ${\displaystyle [G,G].}$ More generally, since conjugation is an isomorphism, any characteristic subgroup is a normal subgroup.

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.

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 ${\displaystyle (123)H=\{(123),(13)\}\ne \{(123),(23)\}=H(123).}$ This illustrates the general fact that any subgroup ${\displaystyle H\le G}$ of index two is normal.

As an example of a normal subgroup within a matrix group, consider the general linear group ${\displaystyle {\mathrm{G}\mathrm{L}}_n(\mathbf{R})}$ of all invertible ${\displaystyle n\times n}$ matrices with real entries under the operation of matrix multiplication and its subgroup ${\displaystyle {\mathrm{S}\mathrm{L}}_n(\mathbf{R})}$ of all ${\displaystyle n\times n}$ matrices of determinant 1 (the special linear group). To see why the subgroup ${\displaystyle {\mathrm{S}\mathrm{L}}_n(\mathbf{R})}$ is normal in ${\displaystyle {\mathrm{G}\mathrm{L}}_n(\mathbf{R})}$, consider any matrix ${\displaystyle X}$ in ${\displaystyle {\mathrm{S}\mathrm{L}}_n(\mathbf{R})}$ 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 ${\displaystyle det(AX{A}^{-1})=det(A)det(X)det(A{)}^{-1}=det(X)=1}$, and so ${\displaystyle AX{A}^{-1}\in {\mathrm{S}\mathrm{L}}_n(\mathbf{R})}$ as well. This means ${\displaystyle {\mathrm{S}\mathrm{L}}_n(\mathbf{R})}$ is closed under conjugation in ${\displaystyle {\mathrm{G}\mathrm{L}}_n(\mathbf{R})}$, so it is a normal subgroup.

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.

The translation group is a normal subgroup of the Euclidean group in any dimension. 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 ${\displaystyle K.}$
• 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. However, a characteristic subgroup of a normal subgroup is normal. A group in which normality is transitive is called a T-group.
• 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 ${\displaystyle G=N⋊H,}$ 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 ${\displaystyle G=M\oplus N.}$
• Normality is preserved under surjective homomorphisms; 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; 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; that is, if ${\displaystyle N_1◃G_1}$ and ${\displaystyle N_2◃G_2,}$ then ${\displaystyle N_1\times N_2◃G_1\times G_2.}$
• 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.
• 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, 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.

Normal subgroups, quotient groups and homomorphisms

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 ${\displaystyle a_1^{\prime }\in a_{1}N,a_2^{\prime }\in a_{2}N.}$ Then there are ${\displaystyle n_1,n_2\in N}$ such that ${\displaystyle a_1^{\prime }=a_1{n}_1,a_2^{\prime }=a_2{n}_2.}$ It follows that $${\displaystyle a_1^{\prime }{a}_2^{\prime }N=a_1{n}_1{a}_2{n}_{2}N=a_1{a}_2{n}_1^{\prime }{n}_{2}N=a_1{a}_{2}N,}$$where we also used the fact that ${\displaystyle N}$ is a normal subgroup, and therefore there is ${\displaystyle n_1^{\prime }\in N}$ such that ${\displaystyle n_1{a}_2=a_2{n}_1^{\prime }.}$ 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 ${\displaystyle eN=N,}$ 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). 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). 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 ${\displaystyle G.}$