In abstract algebra, a normal subgroup (also known as an invariant subgroup or self-conjugate subgroup) is a subgroup that is invariant under conjugation by members of the group of which it is a part. In other words, a subgroup of the group is normal in if and only if for all and The usual notation for this relation is
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 are precisely the kernels of group homomorphisms with domain 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.
A subgroup of a group is called a normal subgroup of if it is invariant under conjugation; that is, the conjugation of an element of by an element of is always in  The usual notation for this relation is
For any subgroup of the following conditions are equivalent to being a normal subgroup of Therefore, any one of them may be taken as the definition.
The image of conjugation of by any element of is a subset of  i.e., for all .
The image of conjugation of by any element of is equal to  i.e., for all .
For all the left and right cosets and are equal.
The sets of left and right cosets of in coincide.
Multiplication in preserves the equivalence relation "is in the same left coset as". That is, for every satisfying and , we have
There exists a group on the set of left cosets of where multiplication of any two left cosets and yields the left coset . (This group is called the quotient group of modulo, denoted .)
Any two elements commute modulo the normal subgroup membership relation. That is, for all if and only if 
For any group the trivial subgroup consisting of just the identity element of is always a normal subgroup of Likewise, itself is always a normal subgroup of (If these are the only normal subgroups, then 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 More generally, since conjugation is an isomorphism, any characteristic subgroup is a normal subgroup.
If is an abelian group then every subgroup of is normal, because More generally, for any group , every subgroup of the center of is normal in . (In the special case that is abelian, the center is all of , 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 of the symmetric group consisting of the identity and both three-cycles. In particular, one can check that every coset of is either equal to itself or is equal to On the other hand, the subgroup is not normal in since  This illustrates the general fact that any subgroup of index two is normal.
As an example of a normal subgroup within a matrix group, consider the general linear group of all invertible matrices with real entries under the operation of matrix multiplication and its subgroup of all matrices of determinant 1 (the special linear group). To see why the subgroup is normal in , consider any matrix in and any invertible matrix . Then using the two important identities and , one has that , and so as well. This means is closed under conjugation in , 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.
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.
If is a normal subgroup of and is a subgroup of containing then is a normal subgroup of 
Every subgroup of index 2 is normal. More generally, a subgroup, of finite index, in contains a subgroup, normal in and of index dividing called the normal core. In particular, if is the smallest prime dividing the order of then every subgroup of index is normal.
The fact that normal subgroups of are precisely the kernels of group homomorphisms defined on 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, and of their intersection and their product are also normal subgroups of
Normal subgroups, quotient groups and homomorphisms
If is a normal subgroup, we can define a multiplication on cosets as follows:
This relation defines a mapping To show that this mapping is well-defined, one needs to prove that the choice of representative elements does not affect the result. To this end, consider some other representative elements Then there are such that It follows that
where we also used the fact that is a normal subgroup, and therefore there is such that 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 There is a natural homomorphism, given by This homomorphism maps into the identity element of which is the coset  that is,
In general, a group homomorphism, sends subgroups of to subgroups of Also, the preimage of any subgroup of is a subgroup of We call the preimage of the trivial group in the kernel of the homomorphism and denote it by As it turns out, the kernel is always normal and the image of is always isomorphic to (the first isomorphism theorem). In fact, this correspondence is a bijection between the set of all quotient groups of and the set of all homomorphic images of (up to isomorphism). It is also easy to see that the kernel of the quotient map, is itself, so the normal subgroups are precisely the kernels of homomorphisms with domain