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The rotations in three dimensions preserving a dodecahedron form a group of order $60$. There are $24$ nonidentity rotations about the six axes through opposite face centres, $20$ about the ten axes through opposite vertices, and $15$ half-turns about the fifteen axes through opposite edge midpoints. Its nonidentity conjugacy classes have sizes $12,12,20,15$: the centralizer of a rotation of order five or three consists of rotations about its oriented axis and has order five or three, respectively, while all edge half-turns are conjugate. A normal subgroup is a union of whole conjugacy classes containing the identity. No proper nontrivial sum of these class sizes plus one divides $60$, proving that it is a simple group. The centralizers of the fifteen involutions are five distinct Klein four-groups, each containing three involutions. Conjugation permutes these five subgroups nontrivially. The kernel is a normal subgroup, so simplicity makes this nontrivial action injective. Its sign map to a group of order two must be trivial, since simplicity would otherwise force an impossible injection of sixty elements into two. Thus its image lies in the alternating group $A_5$, and equality of orders makes the image all of $A_5$.

The orientation-preserving rotational symmetry group of a regular icosahedron has order sixty. Its conjugacy classes have sizes $1,15,20,12,12$: identity; half-turns about opposite-edge axes; rotations through $\pm 2\pi /3$ about opposite-face axes; and rotations through $\pm 2\pi /5$ and $\pm 4\pi /5$ about opposite-vertex axes, respectively. Conjugation carries an axis to its rotated axis and preserves the rotation angle. The transitive group action on edges, faces and vertices gives the indicated classes, with reversal of the axis identifying the two signs. A normal subgroup must contain whole conjugacy classes; none of the proper nonidentity sums of these class sizes is a divisor of sixty. Thus this group is a simple group.

A finite normal subgroup whose order is coprime to its index has a group complement: if $N◃G$ is a Hall subgroup, then $G=NH$ and $N\cap H=1$ for some subgroup $H$. Complement conjugacy holds when either $N$ or $G/N$ is solvable; the existence assertion does not require that extra hypothesis.

For a subgroup $H\le G$ and a normal subgroup $N⊴G$, the map $h\mapsto hN$ from $H$ onto $HN/N$ is a group homomorphism with kernel $H\cap N$. The first isomorphism theorem gives the displayed group isomorphism. The product $HN$ is a subgroup because $N$ is normal. In particular, if $HN=G$ and $H\cap N=1$, then $H\cong G/N$.

A normal subgroup of $A_6$ has order dividing 360 and consists of the identity plus whole conjugacy classes of the alternating group on six letters. If it omits the odd-sized class of size 45, its order is an odd divisor and no nontrivial class sum fits. If it includes that class, the possible class sums not exceeding 180 are $46,86,118,126,136,158,176$, none dividing 360. Every proper subgroup has order at most 180; therefore $A_6$ is a simple group.

For a normal subgroup $N$ of $G$, projection to the quotient group gives a bijection between subgroups of $G$ containing $N$ and subgroups of $G/N$. The maps are $H\mapsto H/N$ and inverse image under projection. Their two composites are identities because each relevant subgroup is a union of cosets of $N$. They preserve inclusions and normality; consequently a composition series in a quotient lifts to a subnormal chain above $N$ with the same composition factors.

Let $G$ be finite and let $H$ have index $p$, the smallest prime divisor of $|G|$. The group action on left cosets gives $\Phi :G\to S_p$, with $\ker \Phi \subseteq H$. The image order divides both $|G|$ and $p!$. By the choice of $p$, the only possible nontrivial prime factor common to these orders is $p$, and $p!$ contains only one factor of $p$. Thus the image order is $1$ or $p$. The action is transitive on $p>1$ cosets, so its image is nontrivial and has order $p$. The first isomorphism theorem gives $|\ker \Phi |=|G|/p=|H|$, hence $\ker \Phi =H$. A kernel of a group homomorphism is a normal subgroup, proving the claim. For example, every subgroup of order $25$ in a group of order $75$ is normal.

A nontrivial subgroup $H$ of $⟨g⟩$ is generated by $g^d$, where $d$ is the least positive exponent with $g^d\in H$. Divide any exponent $k$ with $g^k\in H$ by $d$; its remainder also gives an element in $H$ and must be zero. A quotient group by a normal subgroup is generated by the coset of $g$, because every coset is a power of that coset. The trivial subgroup and quotient are also cyclic.

A subnormal series is a finite chain of subgroups in which each is a normal subgroup of the next. A member need not be normal in the whole group. Requiring the factors to be cyclic groups defines a polycyclic group; allowing finite factors as well defines a poly-(cyclic or finite) group.

Start with a singular disk whose boundary represents a prescribed nontrivial class in a boundary subsurface. Repeatedly lift it through double covering spaces of regular neighbourhoods of its image. The number of distinct image simplexes strictly increases and is bounded by the number of domain simplexes, so the tower ends. At the top, vanishing first mod-two homology forces every boundary component to be a sphere. An embedded disk can then be constructed and projected down the tower. Resolving double curves preserves a boundary class outside a chosen normal subgroup, giving an embedded compression disk at the bottom.

The translation subgroup of a group of Euclidean isometries consists of its translations. It is a normal subgroup, because conjugation by $x\mapsto Ax+b$ takes translation by $v$ to translation by $Av$. Its vectors form an additive subgroup of Euclidean space; for a plane crystallographic group they form a rank-two translation lattice.

The diagonal element $\mathrm{diag}(m,m^{-1})$ conjugates $U(t)$ to $U(m^{2}t)=U(t{)}^{m^2}$. In a finite-dimensional unitary representation, conjugacy therefore makes its eigenvalue multiset invariant under every square power. Square-power rigidity of a finite nonzero spectrum makes the image of every upper unipotent the identity. Its kernel is a normal subgroup; elementary unipotent generators of SL2R then force the entire group representation to be trivial. This algebraic argument does not require continuity.

Let $N$ be a normal subgroup of $G$, and let $\pi :G\to G/N$ be the quotient map of groups. If a group homomorphism $\phi :G\to H$ has group kernel containing $N$, there is exactly one group homomorphism $\bar{\phi }:G/N\to H$ with $\phi =\bar{\phi }\circ \pi$. Its value is forced: $\bar{\phi }(gN)=\phi (g)$. Every measurement by a group homomorphism that ignores $N$ can thus be performed on the quotient group directly. This is the universal property of the construction.