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An abelian quotient is a quotient group $G/N$ that is an abelian group. It exists only for a normal subgroup $N$, and it is abelian exactly when $N$ contains the commutator subgroup. Thus every abelian quotient factors through the abelianization $G/[G,G]$.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. A nontrivial finite p-group has nontrivial centre: in the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.

Every virtually soluble group $K$ has a soluble characteristic subgroup of finite index. First take a soluble normal subgroup $K_0$ of finite index by the subgroup core construction. Choose a soluble normal subgroup $R\supseteq K_0$ maximizing its image size in the finite quotient $K/K_0$. For every soluble normal subgroup $S$, the product $RS$ is soluble, because its quotient by $R$ is a quotient of $S$. Maximality forces $S\le R$. Hence $R$ is the unique largest soluble normal subgroup and is invariant under all automorphisms. This avoids incorrectly intersecting infinitely many conjugates when proving closure under group extensions.

If $H$ is a normal subgroup of a finite group $G$ and $P$ a Sylow subgroup of $H$, thenNormality puts every $g^{-1}Pg$ inside $H$, and Sylow conjugacy there gives $g^{-1}Pg=h^{-1}Ph$. Hence $g{h}^{-1}$ normalizes $P$, giving the factorization. This reduces questions about $G$ to a normaliser and its normal subgroup.

For distinct primes $p,q$, the Sylow theorems force a proper nontrivial normal subgroup in any group of order $p^{2}q$. If $p>q$, the Sylow $p$-subgroup is unique. If $p<q$ and both Sylow counts were nontrivial, the count congruences force $(p,q)=(2,3)$. In order twelve, four Sylow $3$-subgroups exhaust eight nonidentity elements; only three remain for a Sylow $2$-subgroup, making that subgroup unique. Thus the group is not a simple group.

The connected component of the identity in a Lie group is an open normal subgroup. Connected coordinate neighbourhoods prove openness, and conjugation preserves the component of the identity. Every open identity neighbourhood generates it: the generated subgroup is open, its other cosets are open, and connectedness forces it to be the whole component.

If $H$ is a normal subgroup of index $p$ in a group presented by $\mathcal{P}$, the Reidemeister–Schreier theorem produces a presentation $\mathcal{Q}$ withThe preimage of $H$ in the free group has rank $1+p(d-1)$. A relator $w^{p^k}$ produces at most $p$ conjugate relators of weight $p^{-k}$ if its root lies in that preimage. If the root lies outside, $k\ge 1$, and the coset conjugates are redundant up to conjugation; one relator of weight at most $p^{-(k-1)}$ suffices. Thus the total weighted relator cost grows by at most $p$. Convergent infinite sums are handled term by term.

For a normal subgroup $H◃G$, quotient $Q=G/H$, and a $G$-module $M$, the Lyndon–Hochschild–Serre spectral sequence has

A minimal normal subgroup is a nontrivial normal subgroup containing no smaller nontrivial normal subgroup of the ambient group. It need not be a minimal nontrivial subgroup. In a finite soluble group, it is elementary abelian: its derived subgroup must be trivial, a characteristic Sylow subgroup selects one prime, and the characteristic subgroup of elements killed by that prime gives exponent $p$.

The normal closure of a subset $S\subseteq G$ is the smallest normal subgroup containing $S$, equivalently the subgroup generated by all conjugates $gs{g}^{-1}$ with $g\in G$ and $s\in S$.

For a finite Galois extension $L/K$ with group $G$, an intermediate extension $E/K$ is normal exactly when its corresponding subgroup $\mathrm{Gal}(L/E)$ is a normal subgroup of $G$.

If $N=\{1,z\}$ is a normal subgroup of a group $G$, conjugation by any element of $G$ preserves $N$ and fixes $1$. It must therefore fix $z$ as well. Thus $N$ lies in the center of a group. This fact can turn a small normal subgroup into a commuting factor without assuming the whole group is abelian.