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A subgroup $N\le G$ is a normal subgroup when $gN{g}^{-1}=N$ for every $g\in G$. In other words, conjugation by any group element preserves the subgroup as a set, although it may exchange its individual elements. Equivalently, $gN=Ng$ for every $g$: the left and right cosets of $N$ agree. Write $N⊴G$.

Every subgroup of an abelian group is a normal subgroup, because conjugation fixes each element. Normality can also hold in a nonabelian group: in $S_3$, the three even permutations form the alternating group $A_3=\{e,(123),(132)\}$, a normal subgroup as the group kernel of the sign of a permutation. Its two cosets separate the even and odd permutations.

In contrast, $H=\{e,(12)\}$ is a subgroup of $S_3$ but is not a normal subgroup. With function composition of permutations acting rightmost first, choose $g=(123)$. Thenwhich are different cosets. Also $g(12)g^{-1}=(23)\notin H$, directly contradicting invariance under conjugation. Thus a subgroup alone does not suffice for quotient group multiplication.

The practical value of normality is coset multiplication is well-defined exactly for normal subgroups. Each coset can act as one element of a quotient group, with results independent of representatives. Conversely every normal subgroup is the group kernel of its quotient map of groups, so normality also identifies which distinctions a group homomorphism can consistently forget.