Normal subgroup

The condition \(gNg^{-1}=N\) does not ask that each \(n\in N\) stay fixed, only that the whole set be preserved: \(gng^{-1}\) may be another element of \(N\). It is written \(N\trianglelefteq G\). It is equivalent to \(gN=Ng\) for all \(g\), that is, to left and right cosets coinciding; in general \(gN\) and \(Ng\) differ.

Some cases always work: \(\{e\}\) and \(G\) are normal; in an abelian group every subgroup is; every subgroup of index 2 is; the center is; and the kernel of any homomorphism is too (in fact every normal subgroup is the kernel of some homomorphism, as the quotient group will show). A subgroup that is the only one of its order is normal, since conjugation sends it to a subgroup of the same order.

The example worth remembering is \(S_3\): the subgroup \(A_3=\langle(1\,2\,3)\rangle\) is normal, but \(\langle(1\,2)\rangle\) is not, because conjugating by \((1\,3)\) gives \(\langle(2\,3)\rangle\). Normality is not transitive: \(N\trianglelefteq H\) and \(H\trianglelefteq G\) do not imply \(N\trianglelefteq G\), although if \(N\trianglelefteq G\) and \(N\subseteq H\), then \(N\trianglelefteq H\).

A common mistake is to confuse «normal» with «commuting element by element»: \(gNg^{-1}=N\) does not mean \(gn=ng\). That stronger condition describes the center. Also, the intersection of normal subgroups is normal, but the union of two of them is almost never even a subgroup.